LIBRARY OF CONGRES& 

Sfal 1 [njrig|l y a 

Shelf ^/ii. 

UNITED STATES OF AMERICA. 



BY THE SAME AUTHOR. 



A TREATISE 

ON THE 

ADJUSTMENT OF OBSERVATIONS, 

With Applications to Measures of Precision. 



New York: D. VAN NOSTRAND COMPANY. 



TEXT-BOOK 



MECHANIC S, 



With JVuiiebous Examples. 



THOMAS WALLACE WEIGHT, 

Professor in Union College. 



The concrete in which lies always the perenm'aZ.— Carlyle. 




NEW YORK: 
D. VAN NOSTRAND COMPANY, Q 

23 Murray and 27 Warren Streets. 
1890. 



Q? 



Copyright, 1890, 

BY 

Thomas Wallace Wright. 



Robert dkimmond, 

Elect rot yy>< r, 

in .v in; Pearl Street, 

N< w York. 



CONTENTS. 



PAGE 

Introduction, 7 

Chap. I. Motion, - . .9 

II. Force and Motion, . . .30 

III. Dynamics of a Particle, 44 

IV. Statics of a Rigid Body, ..... 86 
V. Friction, 129 

VI. Work and Energy, . . . . . . . 148 

VII. Kinetics of a Rigid Body, . 174 

VIII. Elastic Solids, r 201 

IX. Statics of Fluids, . . . . . . . 212 

X. Kinetics of Fluids, 243 

vii 



PREFACE. 



The science of mechanics, as all other departments of 
physical science, has made great advances in recent years. 
In this book I have endeavored to unfold the principles of 
modern mechanics systematically, and also to call the atten- 
tion of the student to the more useful applications of the 
subject. It has long seemed to me that the more practical 
parts might with advantage be presented to the beginner at 
least as fully as the more abstract and that, too, without 
any sacrifice of scientific precision. The book is so arranged 
that the student, whether intending to make a specialty of 
engineering, physics, or astronomy, can branch out in his 
special direction without difficulty. 

It has been the aim to make the examples as practical as 
possible. Thus while writing the Strong locomotive (No. 
44:4:) was brought out by the L. V. E. E. Co., and the West- 
inghouse air-brake tests (1887) were being made. Many 
problems founded on these and other mechanisms have 
been introduced. 

It has been the aim also to make the examples typical. 
Instead of making them mere numerical illustrations of 
formulas, the idea has been to encourage independent 
thought. In many cases different methods of solution have 
been indicated to encourage the student. to trust rather to 
an independent investigation than to an answer so called. 

Considerable attention has been paid to the graphical 
method of solution, guarding, however, against making it a 
fk complicated weapon with which one can attack all sorts 
of problems which are more easily solved in other ways." 

In several applications of the subject approximate for- 
mulas are of the utmost importance, and many such for- 

v 



mulas are here developed. In all cases the rigorous formula 
has been given first, and the approximate deduced from it. 
In this way the degree of approximation can be estimated. 

As regards the nomenclature of mechanics, I have en- 
deavored to be modern and at the same time conservative. 
One or two terms have been introduced, the better to illus- 
trate the second law of motion and the appropriateness of 
the expression moment of inertia. The words weight and 
pound have been used in the double sense employed in 
ordinary life. No confusion need arise from this, as the 
context is always sufficient to show the sense intended. 
(See p. 40.) 

Other departures from the traditional treatment will be 
noticed. Thus the usual chapter on the Mechanical Powers 
has been omitted, though the " Powers" themselves have 
been discussed in their proper places. All dynamical 
equations have been expressed in terms of the absolute 
units. By doing this it has been possible to ignore the ex- 
pression iv = nig, the source of so much confusion in 
dynamics. 

The Calculus has been used in all cases where its use is 
attended with marked advantage. In the earlier chapters 
two parallel courses are given, one with and one without 
Calculus. This I think an advantage in that it shows the 
student the oneness of what are called elementary and 
analytical mechanics. If thought advisable, a course may 
easily be selected into which no Calculus enters. 

A few historical notes have been interspersed as tending 
to add a living interest. The use of the symbol / as the 
sign of division has added to the compact form of the hook 
by preventing spacing about formulas. 

I have to thank several friends for assistance rendered, 
particularly Prof. Klein of Lehigh, Pres. Staley of Case, 
and Prof. Ziwet of Michigan. 

T. W. W. 

Schenectady, N. Y., June, 1690. 



MECHANICS. 



INTRODUCTION. 

1. If a body occupies a certain position at one time and 
another position at another time we say that the change of 
position must be due to the operation of some cause which 
we call force. If the form of the body has undergone a 
change, we say that the change of form is due to the action 
of force. The fundamental idea of force is derived from 
the use of our muscles, and to muscular effort or to any 
cause producing a like effect we give the name force. 
Muscular effort is exerted .as a push or a pull or some com- 
bination of them. We will therefore for the present con- 
sider force in the sense of a pus7i or a pull no matter by 
what exerted. The science which treats of the different 
effects of force on bodies is called Mechanics.* 

2. Bodies exist in various states which may roughly be 
classed under two general divisions, the solid and the fluid. 
Experiment shows that the result of the action of a force 
on a solid is very different from the result of the action on 
a liquid. Hence it will tend to clearness and simplicity if 

* " Strictly speaking, the derivation of this word should have 
prevented the use of it as the designation of a pure science. It 
has been, however, employed for a long time in English speech in 
the identical sense that the French attach to Micanique pure or the 
Germans to reitie Mechanik. These terms are employed to denote 
what we should prefer to call abstract dynamics — the pure science 
-which treats of the action of force upon matter, which is correctly 
the science of matter and motion." (Tail.) 

7 



8 OTTRODUCTIOU. 

we study the Mechanics of Solids and the Mechanics of 
Fluids separately. 

Experiment shows, too, that the effects of force are very 
different on different solids and on different fluids. Com- 
pare, for example, iron and putty, molasses and water. 
Hence still more minute subdivisions are necessary for pur- 
poses of study. 

3. On account of the infinite variety of solids and fluids 
it becomes necessary as a first step to seize on some phase 
of the problem applicable to all, and then to discuss the 
special cases in detail. Now every substance in nature 
may be conceived to be composed of an indefinitely great 
number of minute portions or particles so small that any 
external force acting on such a particle will act equally on 
all its parts. By considering the action of forces on a 
single particle we may develop principles applicable to 
bodies in all forms considered as composed of particles. 

Still further. Conceive a particle acted on by forces. 
It will in general move in a path of some kind, either 
straight or curved. The particle is so small that the path 
may be regarded as a geometrical line. If we leave the 
forces acting out of view and consider only the positions of 
the particle in this path at different times, the relations 
between these different positions will be in the nature of a 
geometrical problem. Hence, as every particle is endowed 
with at least some of the properties of the body which it 
goes to make up, we may conceive these properties ab- 
stracted from it, thus converting it as it were into a geo- 
metrical point, and consider the motion of this point only as 
it traces out its path during an assigned time and thus be 
able to study motion in its simplest form. To this idea of 
motion in which neither the nature of the particle moved 
nor of the force acting is considered the name Kinematics 
(= science of motion) is given. Kinematics is therefore 
motion in the abstract, and its principles are developed by 



an extension of the principles of geometry by -the introduc- 
tion of the idea of time. 

4. From this bird's-eye view of the subject it is seen that 
the course to be followed is in general outline something 
of this kind : First geometrical motion, next the action of 
force on a single particle and on a system of two or more 
particles either independent or forming a body, and last 
the modifications in body motion resulting from the pe- 
culiar constitution of the various states of matter solid or 
fluid. 



CHAPTER 1. 
MOTION. 




5. The elements of a motion which characterize it are 
change of position {displacement) w r ith change of time. We 
know the position of a body only by noting its relations to 
other'bodies in its neighborhood. Thus 
the position of a point P situated in the 
plane of the paper is defined with refer- 
ence to a point chosen as the point of 
reference, either by measuring the distance OP and the 
angle XOP made by OP with a y Fig. 2 

known line OX (polar coordinates) 
or by measuring the distances PM, 
MO drawn parallel to two known lines 
OY y OX in the plane (Cartesian co- 
ordinates). Any displacement in the plane of the paper* 
would be measured by the changes in these coordinates. 

The motions of different points may therefore be com- 
pared by comparing the changes in their coordinates that 



* If the displacement does not take place in one plane the poiut is 
--similarly referred to three coordinate axes in space, but we shall not 
at present enter into a discussion of this. 



10 MOTION. 

take place in the same time. To render this possible, it is 
necessary to fix on units of angle, time, and distance. 

The unit of angle is the Degree, or the ninetieth part of 
a right angle. The unit of time is the Second. The unit 
of distance is the Foot or Meter, according as we follow the 
British or the metric system. We shall give both systems, 
as it can be done without confusion, and both are in use in 
the application of mechanical principles. Thus the civil 
engineer uses the British system of the foot and its sub- 
divisions, the physicist the metric system, and the electrical 
engineer both systems. 

For convenience of statement distances measured in one 
direction along a line are considered positive, and distances 
measured in the opposite direction negative. The direc- 
tions are arbitrary, but it is usual to take distances to 
the right +, to the left — ; distance upwards -J- , down- 
wards — . 

6. The successive positions occupied by the point in its 
motion trace its Path. This path may be either a straight 
line or a curve of some kind. As most simple, we shall first 
of all consider motion in a straight line (rectilinear motion). 

MOTION IN A STRAIGHT LINE. 

Motion in a straight path may be uniform, or it may vary 
from time to time. Thus an engine on a straight track 
may run for a time with the same speed, or it may vary its 
speed. 

7. Velocity. — The rate at which a moving point changes 
its position is called its velocity. If the rate of change is 
constant, that is, if equal distances are passed over in equal 
times, the velocity is said to be uniform or constant. Thus 
if the point moves uniformly over a distance * in a time / 
its velocity v is s/t. This may be abbreviated into 

V = s/t. 



VELOCITY. 11 

If we place s = 1, t = 1, we find v = 1; or, the wwi£ o/ veloc- 
ity is the distance described in unit time. The measure 
of a velocity is therefore expressed as so many feet (meters) 
per second. 

Clearing of fractions, we may write 

s = vt, 

or the number of units of distance is equal to the product 
of the number of units of velocity by the number of units 
of time. 

As the point may move in the direction of positive dis- 
tance or in the opposite direction, it is necessary to regard 
the sign of the velocity. It is agreed that the same rule 
of signs shall apply to velocity as to distance. The velocity 
is -(- if the direction of motion is in the + direction, and — 
if in the opposite direction. 

The term speed is sometimes used to denote the magni- 
tude only of a velocity. 

Ex. 1. A point moves uniformly with a velocity of 2 
meters per second. In what time will it pass over a kilo- 
meter? Ans. 8 min.^0 sec. 

2. A passenger sitting in a railroad car counts 41 tele- 
graph poles (distant 100 ft.) passed in one minute: show 
that the train is running at 50 miles an hour. 

3. A man a ft. in height walks along a level street at 
the rate of c miles an hour in a straight line from an electric 
light 1) ft. in height: find the velocity of the end of his 
shadow. Ans. bc/(b — a) miles per hour. 

4. The "limited" trains on the N. Y. C. & H. E, E. E. 
running on parallel tracks pass at a certain place in 6^ 
seconds. If each train has the same velocity, and consists 
of 8 Pullman coaches of 52 ft. 9 in. length, find the rate 
per hour. Ans. 46 miles per hour. 

5. Show that the velocity of a point on the equator aris- 
ing from the earth's rotation on its axis is about 463 meters 
per second. 



12 



MOTION. 





Fig. 3 






a h c rf 




vel. 


:U?TC 






A 


£ 


i c 


: c 


) 



8. Curve of Velocity.— A velocity, being distance per 
second, may be appropriately represented by a straight line 
whose length will show the magnitude of the velocity on 
any assigned scale, and whose direction as indicated by an 
arrow will show the direction of the motion. We may 

hence give a geometrical 
representation of constant 
velocity. Along a straight 
line OX (Fig. 3) lay off 
equal distances OA, AB, 
... to any convenient 
scale (as 10 sec. = 1 in.) 

for as many seconds as the 

motion has continued. Let 
the velocities at the points 0, A, B, ... be represented 
by OJ, Aa, ... to any scale (as 10 ft. = 1 in.). Now 
since the velocity is constant, the lines OY, Aa, . . . are 
equal to one another, and the curve of velocity Yd is a 
straight line parallel to OX, the time line. Also 

Dist. passed over = OYx OB 
= area YB; 

that is, the number of feet (meters) in the distance de- 
scribed would be represented by the number of square feet 
(square meters) m the area of the rectangle YD. 

Ex. Plot a velocity of GO miles an hour on a scale of 11 
inches = 1ft. per second, and also the distance passed over 
m one minute. l 

[For plotting it is convenient to use cross-section paper.] 
9. If a point in motion does not move over equal dis- 
tances in equal intervals of time, the motion is said to be 
variable. Variable motion j 8 thus not uniform i„ ( . ve ry 
part of its path. The velocities (),,, An, Bb at the 

times 0, OA, OB, ... if plotted and their extremities 
V, a, o . . . joined would form a line yab . . . not parallel 



TELOCITY. 



13 



case of uniform motion 



Fig. 4 




to the time line OX, as in the 

(Fig. 4). This line would be 

the Telocity curve. The points » 

A,B, . . . may be conceived to 

be taken so close together that 

the velocities between may be 

considered uniform and the 

figures Oa, Ab . . . formed to timc 

be rectangles. Each rectangle represents a distance passed 

over, and therefore the total distance would be represented 

by the sum of these rectangles, that is, by the area 01. 

Conceive now a point to start from 0, and moving with 
uniform velocity Oz, to describe the same distance s in the 
time t as the point moving with the variable velocity. The 
distance would be represented by the rectangle OJL But 
with the variable velocities it is represented by the figure 01. 
Hence the rectangle M is equal to the figure 01, which can 
happen only when Oz is the average (mean) of the values 
Oy, Aa, Bl), . . . Hence, if we can find the average 
velocity v, or the velocity of a point which, moving uni- 
formly, passes over the same distance in the same time t as 
the point moving with variable velocity, we can find the 
distance s described, from the equation 

s = vt. 

10. A velocity curve may also be constructed by laying 
off OA, OB, . . . along 
the line OX to represent 
the distances passed over, 
and Aa, Bb, ... at right 
angles to OX to represent 
the corresponding veloci- 
ties. The line through the 
points a, b, . . . would represent the curve of velocities. 
A familiar illustration is afforded by the motion of the 




14 MOTION". 

piston of a steam-engine. At the beginning and end of its 
stroke the velocity is zero, at the middle of the stroke the 
velocity is greatest, and it varies from this value to the end 
values. The curve of velocity, if plotted, is found to be of 
a form such as Oabc . . . L in Fig. 5. 

11. In uniform motion the velocity v at any instant is 
equal to s'/t. In variable motion the corresponding relation 
would be ds/dt, since the velocity v may be assumed to be 
uniform for an indefinitely small distance ds, and during 
an indefinitely small time dt. Hence 

ds = vdt. 

The total distance passed over in the time t is found by 
summing the distances ds, that is, by the integration 
(= summation) of the expression vdt ; or 



=f t 'vdt, 



which includes all possible cases. 

If, for example, the velocity v is constant, then 

s = vt + c 

when c is the constant of integration. If we suppose that 
the point starts from rest (or 6 = when t — 0) we have 

s — vt, 
as found above. 

Ex. 1. A velocity of GO miles an hour is 88 ft. per sec. 

2. A bicycle rider makes m miles in h hours : find his 
average velocity. Ans. 22m /lbh ft. per sec. 

3. A train runs 29 miles for 2 hours, 30 miles for 3 hours, 
and 32 miles for 1 hour : find its average velocity. 

Am. 30 m. an hour. 

4. The velocity of a particle in a rectilinear path varies 



ACCELERATION. 15 

as it"! distance from the starting-point (= cs): find its posi- 
tion at the end of a time t. Ans. s = e ct , where e is the 
base of the natural system of logarithms. 

12. Acceleration. — In variable motion the rate of change 
of velocity is called the acceleration. If the velocity changes 
uniformly with the time, the motion is said to be uniformly 
accelerated. The measure of an acceleration is the change 
of velocity in unit time. Thus, if a point moves with a 
velocity of 1 ft. per sec, 3 ft. per sec, 5 ft. per sec, etc., 
in consecutive seconds, the acceleration is 2 ft. per sec. in 
every second of the motion. Hence acceleration is distance- 
per-second per second, the unit being 1 foot (or 1 meter) 
per-second per second. 

This method of expressing acceleration, though strictly 
correct, is cumbersome. If it is understood that both 
velocity and rate of change of velocity are referred to the 
same unit of time, it will be sufficient if we express accel- 
eration as distance per second. This is the more ordinary 
method, and with the above understanding can cause no 
confusion.* 

If the velocities of the point had been 5 ft. per sec, 3 ft. 
per sec, . . . the change of velocity is numerically the same 
as before, but in the opposite sense to that of the original 
velocity. The nature of an acceleration may therefore be 
indicated by the signs -\- and — , as the change of velocity is 
in the same sense or in the opposite sense to that of the 
original velocity. Hence an acceleration is -j- if the velocity 
increases algebraically, and — if it decreases algebraically. 
To a negative acceleration the name retardation is some- 
times given. 

13. Just as rate of change of position or velocity is uni- 
form or variable, so may rate of change of velocity or accel- 

* A single word for unit-acceleration is needed, and oue also for 
ynit-velocity. 



16 MOTION. 

eration be uniform or variable. In uniformly accelerated 
motion, the rate of change of velocity being con slant, if u 
denotes the initial velocity, and v the final velocity, the 
change in the whole interval of time / being v — u, the 
change per second or the acceleration a is (v — u)/L and 
therefore 

v — u -\- at (1) 

Also, since the rate of change of velocity is constant from 
beginning to end, the average velocity occurs half-way ; 
that is, 

average vel. = initial vel. -\- \ change of vel. 
= u -f i(v — u) 
= «« + r). 
Hence the distance s passed over is found from 

8 = \{%l + V) X t 

= ut+iaf, fromeq. (1). (2) 

The two equations (1) and (2) contain relations between 
the quantities involved which are independent of one an- 
other. Other relations may be deduced from them which 
are convenient, but which contain no new principle. Thus, 
eliminating t, we find 

v 2 = if -f 2 as, 

a useful formula. 

If the point had started from rest, then u = 0, and the 
equations become 

v = at, v u = 2 ?, 

S = b'/\ s — \rt. 



ACCELERATION. 17 

14. If the acceleration is variable, the relation corre- 
sponding to a — (v — u)/t would be a = dv/dt, since the 
acceleration maybe assumed to be uniform during an indefin- 
itely small change of velocity civ in the indefinitely small 
time clt. Putting v — ds/dt, we have 

a = d 2 s/dt\ 

The formulas found above follow at once from this equa- 
tion. Thus, let a particle start from a point with a 
velocity u : it is required to find its velocity v and distance 
s from at the end of a time t, the acceleratiou being 
constant. Here 

cVs/cW == a. 
Integrating, 

els/ clt = at -|- c, 

c being the constant of integration. But when t = 0, ds/dt 
or v = u; hence 

v = ds/dt — at -f- u, 

which gives the velocity at the end of the time t. 
Integrating a second time, 

s = ±atf -f tft, 

since when t = 0, ,9 = 0, and therefore the constant of inte- 
gration is 0. 

15. Curve of Acceleration. — We have seen how to con- 
struct the curve of velocity of a point in motion, either by 
taking the times as abscissas or the distances as abscissas. 
Both are convenient at times. We proceed now to show 



18 



MOTION. 




how to construct the curve of acceleration when the curve 
of velocity has been plotted. 

First take the case of a motion in which the times are 
plotted as abscissas. Let the velocity change uniformly 
from u to v in the time t. The rate of change of velocity 
being constant, the velocity curve ab . . . of Fig. 4 be- 
comes a straight line, as in 
Fig. 6. The acceleration 
a— (v — u)/t is represented 
in the figure by tan 6, that 
is, by the tangent of incli- 
nation of the line ab . . . to 
OX. Hence if the distances 
OA, AB, . . . represent 
one second, the accelera- 
tions measured on the 
velocity scale would be represented from second to second by 
aa lf bb Jf ... all of which are equal to one another. If, there- 
fore, a line be drawn parallel to OX, and at a distance a 
from it measured on the velocity scale, it will represent the 
curve of acceleration. 

If the rate of velocity is not constant, so that the curve 
of velocity is curvilinear, then since a = dv/dt, the accel- 
eration at a point P (whose co- 
ordinates are v, t) is represented 
by tan 6 when 6 is the inclina- 
tion of the tangent at P to OX. 
Hence, to plot the acceleration 
curve, draw tangents to the 
velocity curve from second to 
second, and lay off as the ordi- 
nates the rise or fall of the tan- 
gent measured on the velocity scale, 
be plotted from point to point. 




The curve may thus 



CURVE OE ACCELERATION. 



19 



Next consider the case in which the distances are plotted 
as abscissas: for exam- 
ple, the curve of accel- 
eration of the piston of 
a steam-engine, i From 
any point P in the 
velocity curve (Art. 10), 
let fall PM perpendicu- 
lar to OX, and draw 
PQ at right angles to 
the tangent at P. Then 
dv 




a = dv/dt = 



ds 



X^ = tan#x PM=MQ. 



Prolong PM to A, so that M A = MQ, and A is a point on 
the required curve. Similarly other points may be found. 
Hence the curve of acceleration is the line AB. 

Comparing the curves of velocity and acceleration in this 
case, we see that when the velocity is zero the acceleration 
is a maximum. As the velocity increases the acceleration 
decreases; when the velocity is greatest the acceleration is 
zero, and at that point it changes sign. 

Ex. 1. If the acceleration is constant, prove that the 
velocity curve is a parabola. 

2. If the acceleration increases uniformly, prove that the 
velocity curve is a straight line. 

16. The relations found for the motion of a particle 
hold also for the motion of a body, provided its motion is 
a motion of translation ; that is, if all the particles of the 
body describe paths that are precisely alike. We therefore 
determine the translation of a body by determining the 
translation of a single particle of the body. The kine- 
matics of body translation is therefore the kinematics of a 
point. 

Ex. 1. In 5 seconds the velocity of a point changes from 
200 ft. to 100 ft. per sec. Find the acceleration. 

Ans. a = - 20. 



20 MOTIOX. 

2. The velocity of a point changes from 20 ft. to 10 ft. 
per second in passing over 75 ft. Find the acceleration 
and time of motion. Ans. a = — 2, t = 5 sec. 

Draw a figure illustrating the motion. 

3. A point starts from rest. Show that the accel. if const, 
is equal to twice the distance described in the first second. 

4. A point describes 1G0 ft. in the first two seconds of 
its motion, and 50 ft. in the next second. When will it 
come to rest? When has it a velocity of 20 ft. per second ? 
When of — 20 ft. per second ? Ans. 5 sec; 4 sec; 6 sec 

5. A point starts with a velocity u and under a constant 
acceleration — a. Show that it will come to rest in u/a 
sec, after describing a distance u 2 /2a. 

6. In the Westinghouse air-brake trials (1887) on the 
P. R. R., a train of 50 freight cars running at 36 miles an 
hour was stopped in 593.5 ft. Find the acceleration of the 
brake. Ans. a = — 2.3 ft. per sec 

7. A point starts from a position A with a velocity u ; to 
find its velocity and its distance from A at the end of a 
time t, the acceleration being proportional to the time 
(= ct) and in the direction of the velocity u. 

Ans. v = u -f Ui' 2 , s = ut + \cf. 

17. Composition of Motions. — Thus far we have consid- 
ered the case of a point which has received a displacement 
in a single direction. But a point may receive several dis- 
placements at the same time either in the same direction 
or in different directions. 

As only one single path results, these displacements must 
combine into a single displacement to which the path is 
due. To this single displacement the name of Resultant 
Displacement is given, and to the separate displacements 
the name of Components. 

Thus, conceive a point P in motion along a straight line 
AB (as a ring sliding along a wire), and 
that at the same time the line is also . p 3 

moved. If the displacement of the ' 

line is in the same direction as that of the point, the point 



COMPOSITION OF VELOCITIES. 21 

receives two simultaneous displacements or a single total 
displacement equal to their sum, 
if in the opposite direction to their / F ' 9 ' ,0 » 

difference. / / 

If the line is moved in the direc- cL ^ 

tion A C while the point is moved / / 

along AB, the point has two simul- / / 

taneous motions — one along AB i q § 

and one along A C. The law of the 

displacements being known, the position of the point may 
be found at the end of any assigned time. For the motion 
along AB alone would carry it a distance AQ in this time. 
But the motion along ^IC'has carried the line AB to the 
parallel position CD, and P will therefore be the position 
of the point at the end of the motion. Hence the final 
position P is the opposite angular point to the initial 
position A of the parallelogram CQ, constructed on the 
lines A Q, A C, representing the distances due to the single 
motions. This proposition is called the Parallelogram of 
Displacements. 

18. ISTow velocities and accelerations being quantities 
having direction and magnitude, and capable of being 
represented by finite straight lines, 
may be treated in a manner analogous vj F,g * " 
to displacements. This we proceed to 
do. 

(a) Composition of Constant Veloci- 
ties. — Suppose a point has two simul- 
taneous constant velocities u, v in the 
directions OX, OY not in the same 
line. 

At the end of the first second the velocity along OX, if 
acting alone, would carry it to a l3 where Oa 1 = u; the 
velocity along OY, if acting alone, would carry it to c lf 
where 0c x = r; when both act, it arrives at d l9 the opposite 




22 MOTION. 

angle of the parallelogram Od 1 to 0. Similarly, at the end 
of the second second it arrives at d, 2 , where Oa 2 — 2u s 
Oc^ = 2v; and so on. The path is thus some line passing 
through 0, d 1 , cl 2 , . . . To find its form we notice that 

C& = 2c 1 d 1 , Oc, = 20c lf 
.'. Oc x : c 1 cl l = Oc 2 : c^l^; 

and hence Od x d 2 . . . must he a straight line. 

Hence the path of the point is along the diagonal 0d 2 ; 
and since the distance Od x passed over in the first second 
is equal to the distances rf, d 2 , . . . passed over in the sec- 
ond . . . seconds, the resultant velocity is constant, and 
is represented by the line Od x d 2 .... This proposition is 
called the Parallelogram of Velocities. 

(b) Composition of Constant Accelerations, The same 
method of reasoning as in (a) may he applied to the com- 
bination of two constant accelerations. The statement is 
Fi g . 12 this: If a point experiences simul- 

a „ b tenuously tivo constant accelerations, 

I ^<*fc / represented in magnitude and direc- 

I ^^ / Hon by two straight lines AB, AC, 

I ^s^/ the resultant motion has a constant 

c D acceleration, which is represented in 

magnitude and direction by the concurrent diagonal AD 
of the parallelogram AD. This is the Parallelogram of 
Accelerations. 

Ex. 1. The velocity along AB is 9 ft. per sec. and along 
A C 12 ft. per sec. If the angle BA C = 90°, find the re- 
sultant velocity. 

[Solution. — Draw AB, AC at right angles. Plot on a 
scale of 12 ft. = 1 in. Then AB = f in., AC = 1 in. 
Complete the parallelogram A BCD. Scale off AD. It 
measures 1J in. Hence resultant velocity = 1£ X 12 
= 15 ft. per sec] 

2. If the angle BAC = G0°, find the resultant. 



RESOLUTION OF MOTIONS. 23 

3. A ball moving with a velocity of 15 ft. per sec. is 
struck so as to move off at right angles with a velocity of 
20 ft. per sec. : find the velocity given to it. 

Ans. 25 ft. per sec. 

19. The above examples having been solved from draw- 
ings made, that is, graphically, solve them a second time 
by computation. 

Ex. 1. This is the same as finding the hypotenuse of a 
right-angled triangle of which the sides are given. Hence, 

AD = VAB 2 + AC* = V9 2 + 12 2 = 15, as before. 

2. This is the same as finding the third side of a triangle 
of which the other two sides and the contained angle are 
given. Hence 



AD = VAB* + AC + 2AB X viG'cos 60°. 

20. Resolution of Motions. Conversely, a velocity or an 
acceleration represented by AD may be broken up or re- 
solved into two components AB, AC, being the adjacent 
sides of the parallelogram constructed on AD as diagonal. 
This may be done in an indefinitely great number of ways, 
as an indefinitely great number of c n s . 13 D 

parallelograms may be constructed on ™^1 

the same diagonal. Other conditions | ^s^ 

must be added to render the problem § ^^ 
determinate. a comp. b 

Suppose, for example, that the components of AD are to 
be at right angles and the angle BAD (= 6) is given. Then 
GAD is known, and the components AB, AC can be plot- 
ted. They are thus determined graphically. 

Since BD — AC, it is evident that the magnitudes of 
the components of AD could be represented by the two 
sides AB, BD of the triangle ABD. Their values may be 
found by solving the triangle ABD trigonometrically. 



24 MOTION. 

Ex. 1. A ship is sailing N. 30° E. at 8 miles an hour: find 
its easterly velocity and its northerly velocity. 

Ans. 4 m., 4 Vd m. per hour. 

2. Find the vertical velocity of a train when moving up 
a \</o gradient at 30 in. an hour. Ans. 0.3 m. an hour. 

3. Solve (1) and ('I) by computing the values required. 

4. Show that the components of a velocity v in two 
directions, making angles of 30°, G0° with it, are v/2, 
vV3/2. 

21. If a particle is at the point x, y at a time t, and if 
ds/dt, dx/dt, dy/dt are the velocity of the particle and its 
components parallel to the axes of X, Y respectively, and 
6 the angle which the direction of motion makes with the 
axes of x, then 

dx ds a du ds . n 



©•=(S)'+(i)' 



Similarly, if d*s/df is the acceleration of the particle in 
its path and the corresponding accelerations parallel to the 
axes are d 2 x/dt 2 , d 2 y/dt'\ then if is the angle which the 
direction of motion makes with the axes of x, 

d 2 x d*s n d 2 y d 2 s . n 

_ = _CO S 0, _j. = __ Hn 0, 



d 2 x 
df 


d*s 
- dt' ( 




\df 



v WW ^ WW ■ 



MOTION IN A CUKVB. 

22. A point P in motion instead of proceeding in the 
same direction may be continually changing the direction 



MOTION IN A CURVE. 



25 



Fig. 14 




Fig. 15 



of its motion so that the path is a curve. The direction of 
motion of P at any point A 
of a curvilinear path ABC 
being the line joining that 
point to the consecutive point 
in the path, will be in the di- 
rection of the tangent A A 1 to 
the curve at A. Similarly, at 
B the direction of motion is 
along the tangent BB X , so that 
in moving from A to B the 
direction of motion has changed from A A. to BB 1} or 
through the angle A X CB X . 

From any point draw a line Oa 
to represent the velocity at A in 
magnitude and. direction, and sup- 
pose a to move so that the line join- 
ing it to will represent in magni- 
tude and direction the successive 
velocities of P as it moves in its 
path. Then a will trace out a continuous curve. 

Consider two positions A and B of P in its path at 
which the velocities are Oa, Ob. The velocity Oa combined 
with the change of velocity between A and B must give the 
velocity Ob at B. But by completing the parallelogram 
Oabd we see that Oa, Od combined will give Ob. Hence 
the change of velocity is represented by Od or by its equal 
ab in the triangle Oab. Now if A and B are indefinitely 
near each other, Oa and Ob represent the values of the 
velocity of P at A and at the next point in the path. 
Hence ab represents the instantaneous change of velocity 
or the acceleration in the path at A. Hence the velocity 
in the curve ab . . . represents the acceleration in the orig- 
Jnal path AB . . . Also the direction of ab is the tangent 
at a. Hence at any point in abc the tangent is parallel to 




26 



the direction of the acceleration at the corresponding point 
of the path ABC. To the curve abc the name of Hodo- 
graph* is given. 

23. For example, suppose a point to move with a uniform 
velocity v in a circular path of radius r. Draw Oa, Ob, Oc, 
F '9- ' 6 ... to represent the velocities 

'&X,A, B, C, . . . in magnitude 
and direction. Each of these 
velocities being equal to v, the 
points a, b, c . . . will lie on a 
circle of centre 0. Also the 
velocity in the circle ABO, 
. . . being uniform, the arcs 
AH, BO, . . . described in 
equal times are equal, and therefore the angles A OB, BOO, 
... are equal. Hence the angles a Ob, bOc, . . . are 
equal, and the arcs ab, be, . . . described in equal times are 
equal, or the hodograph is described with uniform velocitv. 
Hence the hodograph of a point moving with uniform 
velocity in a circle is a circle described with uniform 
velocity. 

_ Ex. 1. A point moves with uniform velocity in a straight 
line: find the hodograph. Ans. A point. 

2. A point moves with uniformly accelerated motion in a 
straight line : find the hodograph. Ans. A straight line. 

3. Show that the direction of motion of any point /; on 
the rim of a wheel running 




with velocity v on a straight 
track is perpendicular to AB 
when A is the point of the 
wheel in contact with the 
track at the instant consid- 
ered. 

[For B has two velocities 
each = v, one along the tan- 
gent BE nM the other along 



Fig. 17 




* By Sir VV. R. Hamilton (1805-1865), 



RELATIVE MOTION. 27 

BD parallel to the path of C. The resultant BG bisects 
the angle EBD. .-. ABG = 90°. The resultant velocity 
= 2v cos 6/2.] 
The solution also follows at once from Art. 145. 

RELATIVE MOTION". 

24. The motion of a point P has been defined by its 
change of position with reference to another point re- 
garded as fixed. This gave the absolute motion of P. 
But if the point is also in motion, or has an absolute 
motion with respect to a third point, the motion of P is no 
longer said to be absolute, but relative. This is really the 
case of bodies in nature, as no point in space is known to 
be fixed absolutely. Still, for the purpose considered, an 
assumed point may be regarded as 'fixed, and in this sense 
motion is said to be absolute. 

The problem of relative motion is really an example of 
the couiposition of motions. Suppose, for example, a body 
A to move with an absolute velocity u, and B with an ab- 
solute velocity v, both being referred to the same fixed 
point 0, and in the same straight line; it is required to 
find the relative velocity of A and B. Conceive both A 
and B to move in a medium which itself moves with a 
velocity v, but in the opposite direction to the motion of 
B. Then B is at rest with reference to the fixed point 0. 
But as the motion of the medium affects A and B alike, 
their relative motion is unchanged. Hence as a velocity v 
has been imparted to A, the velocity of A relative to B 
will be u — v if both were originally moving in the same 
direction and u + v if in opposite directions. 

As an illustration, take two men A and B walking on a 
boat's deck from bow to stern, and that the velocity of the 
boat is equal to the velocity of B. Then B is at rest rela- 
tive to the shore, and the motion of A relative to B is the 
same as if the boat were at rest. 



28 MOTION". 

Ex. 1. Interpret the motion when u — v is +, when — . 

2. If the motions are parallel instead of being in the 
same straight line, are the relative velocities the same as 
before ? 

3. Two trains at a depot are on parallel tracks. Why is 
it difficult for a passenger to tell whether his own or the 
other is in motion ? 

[It is not to be wondered at that it took mankind so long 
to discover that the earth moves round the sun.] 

25. Consider next when the velocities u, v of the two 
points A, B are not in the same straight line. Suppose 
Fig. is the lines OX, OY to represent these 

velocities in magnitude and direc- 
tion. As before, conceive the mo- 
tion to take place in a medium 
~p* x which itself moves with a velocity 
v, but in the opposite direction to 
B, and represented by OZ. The 
V point B is now at rest. The ve- 

locity of A is the resultant of the velocities OZ, OX, that 
is, is equal to the diagonal W, which therefore represents 
the velocity of A relative to B. 

The three velocities are represented by the sides of the 
triangle OXW, the directions being indicated by arrows. 
Hence, if in a triangle one side OX represent the velocity 
of A, XW a velocity equal to and opposite that of B, 
and OX, X W are in the same sense around the triangle, 
the third side OIF taken in the opposite sense around the 
triangle will represent the velocity of A relative to B. 

Conversely, if the absolute velocity of A and the relative 
velocity of A to B were given, we should have from the 
same triangle XW to represent the absolute velocity of B, 
but in the contrary sense. 

Any mechanism may be employed to illustrate relative 
motion by putting a sheet of paper on one of its moving 
pieces, and a pencil on another moving piece, when the 




KELATIVE MOTIOST. 29 

curve traced by the pencil on the paper will represent the 
relative motion of the two pieces. 

Ex. 1. Two vessels start at the same time from the 
same harbor, one sailing east at 12 miles an hour, the other 
south at 9 miles an hour : find the velocity of one relative 
to the other. Ans. 15 m. an hour. 

2. Two railroad tracks intersect at 60°, and two trains 
start at the same instant from the junction at 30 miles an 
hour each : find their relative velocity in magnitude and 
direction. 

3. Two bodies A, B move with velocities u, v inclined at 
*an angle 0: show that the velocity of B with respect to 

A is V u*-\- v 2 — 2uv cos 6, and inclined at an angle 
tan - 1 v sin 6 / (v cos 6 — u) to the direction of A. 

4. Two railroad tracks intersect at 90°. To a passenger 
in one train travelling at the rate of 32 miles an hour the 
other seems to have a velocity of 40 miles an hour: find its 
absolute velocity. A ns. 24 miles an hour. 

5. A boat is propelled at 12 miles an hour across a 
stream flowing at 5 miles an hour, in a direction perpen- 
dicular to the current: find the velocity of the boat with 
reference to the bottom of the channel. 

Ans. 13 miles an hour, up stream. 

6. A man travelling eastward in a wind apparently from 
the north, doubles his speed when the wind appears to 
blow from the northeast. Show that the wind is really 
southeast, and blowing with a velocity of 4 V% miles an 
hour. 



CHAPTER II. 
FORCE AND MOTION. 

26. Common experience shows that to put a body in mo- 
tion or to stop it if in motion requires a certain muscular « 
effort (push or pull) or some equivalent. To this effort or 
its equivalent we give, as already stated, the name force. 

Hitherto we have considered motion apart from the body 
moved and apart also from the force acting. This was an 
ideal case, and gave us the geometrical side of the question 
(Kinematics). In an actual case we must consider the motion 
with reference to the body moved and to the force acting 
as well. This gives us the physical side, and is known as 
Dynamics. In doing this and to lay the foundation for an ' 
extension of the kinematical treatment of motion to dynam- 
ical questions, we are compelled to call in the aid of ex- 
periment. 

The science of dynamics rests upon three principles or 
laws, known as Newton's laws of motion.* Before stating 
them, certain rude experiments will be indicated which are 
sufficient to suggest the laws but not to establish their 
truth. No direct proof is possible. The proof is indirect, 
and is made in this way. Assume the laws true, and cer- 
tain consequences follow which can be tested experiment- 
ally. This has been done in so many ways and by so many 

*•'. . . Though Newton's laws of motion are a much clearer and 
more general statement of the grounds of Mechanics than had ap- 
peared before his time, they do not involve any doctrines which had • 
not been previously stated or taken for granted by other mathema- 
ticians. " — Whewell. 

30 



STRESS. 31 

independent observers, particularly in astronomical work, 
that Ave are justified in accepting them as true. So com- 
plete have been the tests and so firm is the conviction of 
the truth of the laws, that results deduced from them which 
cannot be verified in any way are accepted without hesita- 
tion. 

27. Stress. — In order to exert force the agent acting must 
meet a resistance. Thus the hand in motion does not exert 
force until it meets some object. The object reacts on the 
hand. Press the table and the table will press the hand. 
Force is always & mutual action: in other words, forces are 
never single, but act in pairs — one the action and the other 
the reaction. This pair of actions is known as a Stress. 
If it is of the nature of a push, as in cracking a nut, the 
name compression (or pressure) is given to it ; if of the na- 
ture of a pull, as in breaking a string, the name tension is 
given. 

A stress can never occur except between two bodies or 
two parts of the same body, and is always exerted over the 
surfaces coming in contact. If we divide the total stress 
by the area of this surface we obtain the stress per unit 
area, as per sq. in., for instance. Thus we are accustomed 
to speak of the stress of steam in a boiler as pressure per 
sq. in. Or we may consider stress without reference to 
area, looking to the total magnitude only, and consider it 
acting at a point. 

In some cases the relation between the action of the agent 
and the reaction of the resistance is sufficiently evident. 
Thus if one body rests upon another, it will be granted that 
the pressure exerted by the upper is equal to the counter- 
pressure exerted by the lower : if a horse hauls a canal-boat 
to which he is attached by a rope, the pull of the rope on 
the horse is equal to its pull on the boat, and so on. But 
when a stone falls from a height it is not evident whether 
the action of the earth on the stone is equal to the action of 



32 FOECE AND MOTION. 

stone on the earth. Nor is the relation evident between the 
actions of a magnet and a piece of iron, nor between bodies 
widely separated, as the earth and moon. 

The results of experiments direct and indirect on the 
mutual actions of bodies are summed up in what is known 
as the law of stress : To every action there exists always a 
reaction equal in •magnitude and. opposite in direction ;. or 
as it may be expressed : The mutual actions of two bodies 
are always equal, and act in opposite directions, 

28. Force being thus always double may be looked at 
from the point of view of the body acted on or of the body 
(as agent) exerting the action. The two components being 
equal and opposite, either will suffice for a numerical meas- 
ure of the magnitude of the force. 

Suppose a body at rest on a level floor. If let alone it 
will remain at rest. If a push is given to it (= force ap- 
plied) so as to cause it to move along the floor, it will come 
to rest after going a short distance. If the floor is waxed, 
it will go a greater distance for the same push. The 
smoother the surface the farther it goes, and the more 
nearly in a straight line. If the floor were perfectly smooth, 
we can conceive of no reason why the body should not con- 
tinue to move in a straight line forever. 

Now our minds are so constituted that we cannot con- 
ceive of a change occurring without a cause. On the rough 
floor the change of motion is accounted for by the action 
between the floor and the body — an action outside the body. 
But on the smooth floor this outside action is removed. 
Still, in this case, to make the body change its velocity or 
change the direction of motion, some outside action is 
found to be necessary. We say then that the body has 
within itself no power of making any change in its state, 
either of motion or of rest. To this property of inability 
of a body to change its state we give the name Inertia.* It 

* The term inertia or vis inerlice was introduced by Kepler (1571- 
1630) 



is to be regarded as an inherent or characteristic property 
of matter. The law of inertia was enunciated by Newton 
as follows : Every body continues in its state of rest or of 
uniform motion in a straight line except in so far as it is 
compelled by impressed forces to change that state. 

It thus appears that force causes not merely change of 
place in the body moved, but change of velocity as well. 
We may therefore extend the definition from muscular 
effort to this: Force is whatever causes deviation from uni- 
formity or reel Hlinearity of motion in the body acted on. 

29. The force exerted by an agent on a body being one 
component of the stress, the other component, the reaction 
of the body or the " kick against change of motion," de- 
pends on the inertia of the body, and may be called inertia- 
resistance. Being equal to the action of the agent, if Ave 
can measure it we have a measure of the stress, which term 
includes both components. 

We have many familiar illustrations of inertia-resistance. 
Thus, on jumping from a train in motion, on reaching the 
ground we are hurled forward. In jumping on a train in 
motion, a jerk is received. So a sudden change of velocity 
will make itself known to the passengers by a thrust or 
jerk. The intensity of the thrust or jerk depends on the 
difference of velocity of train and passenger, thus showing 
that the inertia-resistance called into play is proportional 
to the instantaneous change of velocity or the acceleration 
a communicated. We may therefore write 

inertia-resistance = ma 

when m is a constant. This constant may be called the 
coefficient of inertia. 

30. The comparison of forces is thus reduced to a com- 
parison of distances and times, and by assuming a unit of 
lorce all forces may be expressed in terms of this unit s< i 
long as we keep to the same body. 



34 FORCE AND MOTION". 

Mass. — But in general the question is not regarding the 
comparison of forces acting on the same body, hut of forces 
acting on different hodies. Suppose two passengers to 
jump on the same train, the larger will receive the greater 
jerk. We explain this by saying that the two bodies are 
of different Mass. "We may therefore compare the masses 
of hodies by communicating the same acceleration to each 
body when the ratio of the masses will he the ratio of the 
coefficients of inertia. A rude measurement would he af- 
forded by having each passenger grasp at the instant of 
jumping on the train a spring-balance: the ratio of the 
pulls would indicate the ratio of the masses. 

We may observe the effects of the same force on blocks 
of the same substance of different sizes. It is found that 
the larger the block the smaller the acceleration imparted 
by the force, that is, the greater the mass. The mass would 
in this case depend on the size of the block, or, as is some- 
times said, the quantity of matter in it. And when we 
come to blocks of different materials we agree conven- 
tionally to ascribe the greater mass to the body to which the 
smaller acceleration is imparted. If the same force gives 
the same acceleration to two different bodies, we say conven- 
tionally that they are of the same mass: and if the accelera- 
tion given to one is n times that given to the other, we Bay 
that the mass of the second is n times that of the first. 
Hence, the masses of bodies are inversely proportional to 
the accelerations imparted to them by the name force. 

31. Unit of Mass. — It follows that masses may be com- 
pared by a measurement of accelerations. The operation 
may be called massing. It enables us to express all masses 
in terms of some one standard mass, and so make a 
quantitative measurement of mass. This one mass we may 
choose for unit-mass, and we may make it what we please 
As in the case of the other fundamental units already 
assumed, the units of distance and time, all that is neees- 



MASS. 35 

sary is that having once chosen it we must be consistent in 
its use. 

The units employed are the British unit, which is a 
certain lump of platinum called a Pound (lb.), and the 
metric unit, which is also a lump of platinum called a 
Kilogram (kg.). The masses of all bodies may now be ex- 
pressed in pounds or kilograms by comparing the accelera- 
tions due to the action of the same force on the unit mass 
and on the bodies in question. Multiples or submultiples 
of the standard units, such as the ton, ounce, gram, etc., 
are often employed as being more convenient in certain 
cases than the standard units themselves. 

32. Unit of Force. — We have seen that forces may be com- 
pared by comparing the accelerations produced in the same 
mass. This enables us to express all forces in terms of a 
unit force. The units of acceleration and mass being 
already defined, the unit force may be conveniently ex- 
pressed in terms of them as the force producing unit- 
acceleration in unit mass.- If we take unit acceleration to 
be one ft. per sec, and unit mass to be one lb., the name 
Poundal is given to unit force; if unit acceleration be one 
cm. per sec. and unit mass be one gram, the name Dyne is 
given to it. 

33. We may find a numerical expression for any force in 
terms of the unit force, say the poundal. For by our 
definition, to impart to a mass of one lb. an acceleration of 
one ft. per sec. requires a force of one poundal, and the 
preceding makes it reasonable to suppose that to impart to 
a mass m an acceleration a would require a force of ma 
poundals. Denoting the force by F, we may write 

F — ma, 
which means that the force which produces an acceleration 
a ft. per sec. in a mass of m lb. is ma poundals. 
*- Now the force F and the inertia-resistance of the body 



36 FOKCE AND MOTION 

being opposite aspects of the same stress, it follows from 
Art. 30 that with this system of units the coefficient of 
inertia is equal to the number of units of mass contained in 
the body, and the inertia-resistance is equal to the mass- 
acceleration ma of the body. 

If now u is the velocity of a body of mass m at the be- 
ginning of time t and v the velocity at the end of this time 
under the action of a constant force F, the acceleration a 
is equal to (v — u)/t (Art. 13). Hence, substituting for 
a in F = ma, we have 

Ft = mv — mu. 

The product Ft is called the impulse of the force during 
the time t, and the product mv or mu is called the mo- 
mentum of the mass m in the direction of the velocity v 
or u. Hence mv — mu is the change of momentum due to 
the impulse. 

We may now appreciate more clearly the meaning of New- 
ton's second law of motion, which states that: The change 
of momentum of a body is numerically equal to the impulse 
which produces it, and is in the same direction. 

34. From the law we have the relation 

Ft — mv — mu, 

as its statement in symbolic form. 

Writing this in the form F= (mv — mu)/t, we see that 
force may be defined as rate of change of momentum. 

Putting the rate of change of velocity (v — u)/t = a, the 
acceleration, we have 

F = ma, 

a convenient form, and which is called the general equation 
of mo I ion. 



Whether the acceleration be uniform or variable, we may 
write (Art. 14) 

a — cVs/clf, 
and hence 

F=md 2 s/dt 2 , 

which is also called the general equation of motion. 

The general equation of motion, which is the algebraic 
statement of the second law of motion, is the connecting 
link between motion and force. It enables us to pass from 
the kinematical properties of motion already laid down to 
questions involving force and mass. It is the link between 
the ideal and the actual, the geometrical and the physical. 

35. From this law we infer, too, that since change of 
momentum per second is proportional to the magnitude 
only of the force acting, this change is the same whether 
the body is at rest or in motion. 

Hence, too, for the same-body force and acceleration are 
simultaneously constant, being connected by the relation 
F— ma; or, as it maybe stated, a constant force constitutes 
constant acceleration to the body acted on by it. 

A variable force maybe considered to consist of a succes- 
sion of constant forces varying in magnitude and direction, 
and acting for indefinitely small intervals. The accelera- 
tions contributed may be considered uniform during these 
intervals, and the total acceleration in any time found by 
summation. 

The law also implies that when two or more forcss act 
on a body at the same time, each force produces an accel- 
eration in its own direction without reference to the others. 
It therefore follows that forces may be combined by the 
rules already laid down for the combination of accelerations 
(Art. IS). 

Ex. 1. Explain why by striking the handle of a hammer 
against a wall the head may be fixed on firmly. 



38 FORCE AND MOTION. 

2. A man stumbling can save himself more easily on land 
than on smooth ice. Explain. 

3. In suburban-passenger traffic the trains must stop 
and start quickly. The boiler and machinery are placed 
over the driving-wheels. Why ? 

4. Show that it necessarily follows from the second law 
of motion that forces can be represented by straight lines. 

5. What are the tests of the equality (1) of two forces, (2) 
of two masses ? 

6. State the parallelogram of momenta. 

7. A man with a hod on his shoulder fulls off a ladder : 
find the pressure on his shoulder during the fall. 

36. Gravitation Measure of Force. — The unit of mass be- 
ing assumed, we have seen how all masses may be expressed 
in terms of the unit by measuring the accelerations contrib- 
uted by equal forces. This is a strictly scientific method 
of measuring mass, and is sufficient. But though easily 
described in general terms, it is difficult of performance in 
practice. Accordingly we give another method more easily 
put in operation. 

It is a fact of common observation, that a body free to 
move falls towards the earth. It acts as if the earth at- 
tracted it. It is assumed as a convenient explanation of 
the observed phenomenon that there exists an attraction 
between the earth and the body, and to this attractive force 
the name force of gravity is given. 

A body free to move if exposed to the action of the force 
of gravity is uniformly accelerated. Experiment* shows 
that at the same place it acts on all bodies in the same way; 
that is, the acceleration g produced by it has no relation to 
the magnitude of the bodies or to the material of which 
they are composed. 

* Drop simultaneously pieces of lead, iron, paper, etc., etc., from 
a shelf iu a glass vessel from which the air has been exhausted. All 
will be observed to strike the bottom of the vessel simultaneously. 
The only force acting is the force of gravity, and since all strike the 
bottom at once, they must have the same acceleration, as each has 
passed over the same distance in the same time. 



GRAVITATION MEASURE OF FORCE. 39 

Experiment shows, too, that the value of g is constant so 
long as we keep to the same place on the earth's surface. 
It varies, however, from place to place.* This is explained 
by the fact that the earth is not a perfect sphere, and is not 
homogeneous in structure. For latitude 45° at sea-level its 
value is 32.2 ft., or 9.81 meters nearly. This may be taken 
as an average value. 

37. "We may make a rough comparison of any force with 
the gravity force of a body in this way. Place the body on 
a spring-balance and note the compression of the spring. 
If the force produces the same compression of the spring 
we say it is equal to the gravity force of the body. 

The unit force is naturally assumed to be equal to the 
gravity force of the unit mass, one lb. or one gram. This 
force causes, as explained above, an acceleration g in one lb. 
But the force causing an acceleration g in one lb. is^/ pound- 
als. Hence the unit force is equivalent to g poundals. As, 
however, it is convenient tojiave a distinct name, the unit 
is called a Pound, so that a force of one pound is equivalent 
to the attractive force between one pound mass and the 
earth, and is equal to g poundals. Hence we may convert 
poundals to pounds by dividing by the value of g at the 
place in question. 

38. The use of the word pound in the double sense of mass 
and force is objectionable, but it is sanctioned by ordinary 
custom, and the context must decide in which sense it is 
used. To aid in making the distinction we shall use for 
mass the symbol lb., and for force the word^ow^.f 

The pound is called the gravitation unit of force, because 
it depends on the force of gravity, which is not constant in 

* "When Halley in 1677 went to the island of St, Helena to observe 
the stars of the southern hemisphere he found his clock lose so much 
that the screw at the bottom of the pendulum did not enable him to 
shorten it sufficiently." 

f Suggested by Supt. Mendenhall of the United States Coast and 
Geodetic Survey. 



40 FORCE AND MOTION. 

value over the earth's surface. The other unit of force, 
the poundal (or dyne), does not involve g, and is known 
as the absolute unit. The first being the older and more 
easily applied, is the unit of daily life; the second being the 
more comprehensive is used in astronomical and electrical 
work, and in precise physical investigations in general. 

39. Weight. — The gravity forces of two bodies of masses 
m, m 1 being as mg poundals to m/j poundals or as m to m l 
are in the ratio of the masses. If the bodies are placed in the 
scale-pans of a common balance and the balance remains in 
the same position, the gravity force of each is the same, and 
the two bodies have equal mass, or, as we say in common 
language, have equal Weight.* The process is called weigh- 
ing, and hence, by assuming a body of standard mass, we may 
express all bodies in terms of this standard. We may there- 
fore, in comparing masses, substitute for the complicated 
process of massing (Art. 31) the simple operation of weigh- 
ing. 

In ordinary language the word weight is used in the 
double sense of mass and force. The original signification 
of the term was what we now call mass, and its extension to 
force was a later development. " The word weight must 
be understood to mean the quantity of the thing as deter- 
mined by the process of weighing against standard weights." 
Thus in buying a barrel of flour we buy the mass that weighs 
19 G pounds. 

The builder and machinist find it necessary to use it in 
the other sense, as in computing the force necessary to 
support a load of given weight (= balancing the gravity 
force of a load), or in computing the stresses in a structure 
designed to support a given load. 

It is unfortunate that the term is ambiguous, but there is 
no help for it — any more than for the ambiguity of the term 

* On the effect of "centrifugal force" on the weight of a body 
see Art. 69, with the examples appended, particularly 9—11 inclusive. 



.NOTE ON UNITS. 41 

pound. In fact the two go together — pound weight (mass) 
originally, and pound weight (force) secondarily. 

To prevent confusion, we shall express all dynamical 
formulas in terms of the absolute units. The passage to the 
gravitation units and the use of the terms weight, pound, 
etc., in the examples can give no trouble if the explanations 
given are kept in mind. The context will always make 
clear the sense intended. 

Ex. 1. How many poundals are equal to [the gravity force 
of] one ton? Ans. 2000 g poundals. 

2. Show that one poundal is equivalent to \ oz. 

3. Would it be advantageous for a merchant to use a 
spring-balance for buying groceries in New York to sell 
in Cuba ? How would a pair of scales answer ? 

40. Note on Units. — The three fundamental units in 
Mechanics are the units of time, distance, and mass. Being 
fundamental, they are arbitrary, and are chosen for con- 
venience, or as the result of circumstances. Their defin- 
itions have already been given ; but we shall here repeat, 
collect, and go into a little more detail. 

The unit of time is the Second. It is derived from ob- 
servations of the earth's rotation. The assumption of this 
unit therefore really amounts to making the motion of the 
earth on its axis the standard motion, and by means of the 
second all motions are tacitly compared with this standard. 

The standards of length are the Yard and the Meter. The 
yard is the distance between two lines on a certain bronze 
bar kept in London, England, when the bar is at the tem- 
perature 62° F. The foot is J of this distance, and the inch 
gJg of the same distance. The meter is the distance between 
the ends of a certain platinum bar kept in Paris, France, 
when at the temperature 0° C. It was intended to be the 
ten-millionth part of the meridian distance between the 
equator and the pole, and is nearly equal to this distance, 
but not exactly. A meter is equal to 39.37 inches, or 3.28 
ft. The centimeter is T -J-y of the meter. 



42 



FORCE AND MOTION. 



The standards of mass are the Pound (lb.) and the Kilo- 
gram. The pound is a certain piece of platinum kept at 
London, England. The ounce is -f-% of the pound. The 
kilogram is a certain piece of platinum kept at Paris, 
France. The gram is r¥ Vo °f the kilogram, and was in- 
tended to be of the same mass (which it is very nearly) as a 
cubic centimeter of water at 3°.9 C, the temperature of 
maximum density of water. 

The units used in any investigation are multiples or sub- 
divisions of the standard units as found most convenient. 
The system of units involving the centimeter, gram, and 
second, with the dyne as unit of force, is called the C. G.S. 
system. It forms a sort of international system, and is being 
largely adopted by physicists, astronomers, and electricians. 
The British (absolute) system of the foot, lb., and second, 
with the poundal as unit of force, is known as the F.P.S. 
system; and the British (gravitation) system of the foot, 
lb., and second, with the pound as unit of force, is the sys- 
tem of every-day life. In engineering work, the inch, ton, 
and minute are very often the units employed. 

The following table of relative magnitudes will be found 
convenient : 



1 lb. = 453.59 grams. 
1 inch = 2.54 centimeters. 
1 foot = 30.48 centimeters. 
1 mile - 1609.33 meters. 



1 gram = 0.0022 lb. 

1 centimeter = 0.3937 inch. 
1 meter = 3.2809 feet. 

1 gram = 15.432 grains. 



41. Dimensions of Units. — All mechanical quantities are 
expressed in terms of some system of units. The two leading 
systems in use in this country, the F.P.S. and the C.G.S., 
have been explained. It is convenient to be able to pass 
rapidly from the one system to the other, or from one Bystem 
to any other. Of course the quantity itself is quite inde- 
pendent of the unit employed to measure it,— just as the 
matter of this book is in no way affected by the size of the 
type used by the printer. 



NOTE ON UNITS. 43 

The fundamental units are those of distance,, time, and 
mass. Let L, T, M denote the magnitudes of these units. 
Then if we say a distance is I units in length, the complete 
symbol representing this would be IL. Usually it is written 
I only, the unit being tacitly assumed. 

All derived units may be expressed in terms of the fun- 
damental units. Thus unit velocity being the Telocity of 
a point which describes unit distance in unit time, we have 

unit vel. = unit dis. / unit time = L/T. 

Similarly, 

unit accel. = unit vel. / unit time = L/T 2 ; 

unit force = unit mass X unit accel. = 21 L/T*; 

unit mom. = unit mass X unit vel. = ML/T; 

and unit impulse being measured by the number of units 

of momentum generated is of the same dimensions as unit 

momentum. 

Ex. 1. How manv dynes-in a poundal? 
[1 poundal = ML/T 2 = lb. X ft./sec. 2 = 453.59 grams X 
30.48 cm./sec. 2 = 13S25.3 dynes.] 

2. Show that the foot per sec. and mile per hour units of 
velocity are as 15 : 22. 

3. Show that one mile per hour is 44.7 cm. per sec. 

4. Reduce a velocity of 100 ft. per min. to cm. per sec. 

5. How many dynes iu [the force of] an imperial pound? 

Ans. 445,000 dynes, about = f megadyne. 

6. Show that [the force of] 1 gram = 981 dynes. 

7. Show that [the force of] 1 grain = 63.58 dynes. 

8. Prove that a pressure of 1 pound per sq. ft. is equiva- 
lent to 479 dynes per sq. cm. 

9. Find the value of g if one minute is taken as the unit 
of time. Jus. 115,200 ft. 

10. Find the unit of time if (j is taken as unity, one ft. 
being the unit of length. Ans. 0.25 V^> sec. 

11. How many dynes in the unit of force if the meter, 
minute, and kilogram are taken as the units of distance, 
time, and mass respectively ? 

Am. 100 x 1000 x GO 2 dynes. 



CHAPTER III. 

DYNAMICS OF A PARTICLE. 

42. Haying considered the geometrical properties of 
motion, and also the methods of measuring mass and force, 
wo are ready to study the motion produced in a body of 
given mass by forces of given magnitude. 

When a body is acted on by forces, experience shows that 
various forms of motion may arise. If all of the component 
particles of the body move through equal distances in the 
same direction, the motion is said to be a motion of Transla- 
tion. The motion of any particle would in this case give 
the motion of the body. In a motion of rotation the par- 
ticles do not move through equal distances in the same 
direction, those nearest the axis of rotation moving the 
shortest distance. If a body consisted of a single particle, 
it would, in its rotation about an axis passing through it, 
remain in the same position. We shall therefore exclude 
rotation, aud be able to study the translation of a body if we 
consider the motion of a single particle only. We may con- 
ceive the whole body concentrated as it were into a single 
particle of equivalent mass. 

43. Composition of Forces. — The number of forces acting 
on a particle may be one or more than one. If we can 
combine the separate forces into an equivalent single force, 
we can reduce all cases to that of the action of a Bingle 
force. The method of combining forces will be our first 
step, and next we shall consider the motion of a particle 
under the action of a force. 

If several forces act on a particle at rest or in motion, the 

44 



COMPOSITION OP FORCES. 45 

accelerations communicated are the same as if each force 
acted separately on the particle at rest. In other words, 
the acceleration produced by a force on a particle is inde- 
pendent of any motion it may have and independent of 
motions produced by other forces acting simultaneously. 
This is involved in the statement of the second law of mo- 
tion, but is repeated and expanded here for greater clear- 
ness. It is sometimes known as the principle of the 
independence of forces. 

44. Representation of Force. — When a force acts on a 
particle its line of action must pass through the particle. 
The force itself contains a certain number of poundals or 
dynes. We may say that the elements of the force are 
three — the geometrical position of the particle acted on, the 
direction of the force, and the number of units it contains 
or its magnitude. It may therefore be represented by a 
straight line AB, the length of AB representing the mag- 
nitude of the force, the direction from A to B the direction 
of the force, and the point A the point of application. 
Each unit of length of A B will represent unit force. But 
the length of the unit is arbitrary. Hence we may plot 
forces to any scale we please, as 1 poundal = 1 inch, 10 
poundals = 1 inch, etc. 

The direction of a force may be indicated by its sign. 
Thus a force of 2 units acting at A towards the right might 
be written -f- 2, an equal force in the Fi , 9 

opposite direction — 2. The choice <. -\ *— 

of signs is arbitrary, and it is only A 

necessary to remember that if one direction is assumed to 
be -(-, the opposite direction must be — . In a diagram the 
direction is conveniently indicated by an arrow-head. 

Ex. On a scale of 100 pounds per in. what force would 
be represented by a line 20 in. long ? Ans. 2000 pounds. 

45. Resultant of Two Forces. — Suppose that F x , F„ are 
two forces which act on a particle A of mass m. The 



46 



DYNAMICS OF A PARTICLE. 




Fig. 21 



accelerations a, , a g produced by these forces are (Art. 34) 
FJm, FJm respectively, and the resultant acceleration a in 
direction and magnitude is repre- 
Fig. 20 sented (Art. 18) by the diagonal of 

the parallelogram of which a l9 a 2 are 
adjacent sides. Hence, if R denotes 
the force which would produce the 
acceleration a, that is, if R = ma, it 
must represent in magnitude and 
direction a force which produces the same acceleration on 
A as F 1 and F 2 . To the force R the name Resultant Force 
is given, and conversely F x , F^ are the Components of R. 

We may represent this graphically. Let AB, A C repre- 
sent the forces F i , F 2 in magnitude 
and direction (scale, say, 1 poundal 
= 1 in.); then AD the concurrent 
diagonal of the parallelogram con- 
structed on AB, AC as adjacent 
sides will represent in magnitude 
and direction the resultant R of the 
two forces on the same scale. This principle is called the 
Parallelogram of Forces.* Hence, instead of finding the 
accelerations due to the forces and combining them into a 
single acceleration, and thence finding the force which would 
produce this acceleration, we may combine the forces them- 
selves directly. 

Ex. 1. If the two forces act in the same straight line, 
find their resultant. Ans. F t ± F 2 . Why the double 
sign ? 

2. Explain the action of the forces by whicli an arrow is 
discharged from a bow. 

3. Explain (by a drawing) the action of the forces by 
which a kite rises in the air. 

4. Show by a drawing that the value of R decreases as 
the angle between the forces increases. 

* The parallelogram of forces was first formulated by Newton 
(1642-1727). 





COMPOSITION OP POECES. 4/ 

5. A satchel is carried by a strap slung over the shoulder : 
show that the longer the strap the less its tension. 

46. Instead of finding the diagonal of the parallelogram 
by a geometrical construction, as 
just explained, we would obtain the 
same result if we computed its value 
from the known values of the sides 
AB, A C and the contained angle 
BAC (= 8). Thus from trigonom- 
etry we have in the triangle ABD 
[note AC=BD; BAC + ABD = 180 c ]. 

AD" = AB 2 + BD 2 - ZAB\ BD cos ABD 
= AB' + AC 1 + 2AB .AC cos BAC, 
or R 2 = F* + F 2 + %F^F % cos 6, 

which gives the magnitude of the resultant. The direction 
may be found by solving the triangle ABD to find the angle 
BAD, The position is known since the force acts at A. 
Hence R is completely determined. 

The special case of the forces acting in directions at right 
angles is important. The parallelogram becomes a rectan- 
gle, and from the figure 

R> = F* + F*, cos BAD = FJR, 

whence both the magnitude and direction of the resultant 
are determined. ♦ 

Ex. 1. If two equal forces F, .Fare at right angles to ona 
another, then R = FV'2. 

2. If two equal forces F, F are inclined at an angle 2 6 y 
then R = 2F cos 8. 

3. Prove that the value of R increases as the angle be^ 
tween the forces diminishes, and vice versa. 

•4. When the angle BAC is 180° the forces are in the 
same straight line, and the formula, if correct, should re- 
duce to the sum or difference of the forces. Examine, and 
see if it does. 



48 



DYNAMICS OF A PARTICLE. 




5. Two equal forces .Pact at an angle of 60°: find their 
resultant. 

6. Find the resultant of two equal forces, each of 10 
pounds, acting at an angle of 30°. Ans. 19.3 pounds. 

47. Consider in Fig. 22 the manner in which the result- 
ant is formed. The forces F., F 2 drawn to scale are rep- 
resented by the lines AB, AC. From B the hue BD is 
drawn parallel to AC, and from C the line CD is drawn 
parallel to AB. The diagonal AD of the parallelogram 

represents the resultant in magnitude, 
direction, and position. 

This construction is equivalent to 
that shown in Fig. 23. Plot the 
forces F l , F^ as before. From B draw 
BD parallel and equal to A C. Join 
AD, which represents the resultant. 
Still better, by breaking the figure into two parts (Figs. 
24,25). Let F l9 F % be Rg ^ 

the forces acting at 0. 
From any point A draw 
AB to scale equal and 
parallel to F l . From B 
draw BD to scale equal 
and parallel to F 3 . Join AD, which will represent the 
resultant in magnitude and direction. To find its position: 
We know that it must pass through 0, and hence if through 
we draw a line equal and parallel to AD, Ave have R in 
magnitude, direction, and position. We have therefore a 
Force Diagram (Fig. 24) and a Construction Diagram purely 
geometrical (Fig. 25). In Fig. 23 the two overlap, and in 
simple cases there is no confusion in their being so drawn; 
but in complicated cases it is better to keep them separate, 
as we shall see later. 

48. Resultant of more than two Forces. — Let F lt F 9 ,F 9 , 

F A represent forces acting on a particle at : it is required 
to find their resultant. 




COMPOSITION OF FORCES. 



49 



Fig. 26 



Fig. 27 




Fig. 28 



Following the method of Art. 47, from a point A we draw 
AB equal and parallel to 
F l , BC equal and parallel • 
to F^ , CD equal and par- 
allel to F 3 , DE equal and 
parallel to F i . Join A E, 
which will represent the 
resultant in magnitude 
and direction. In the force diagram draw R equal and 
parallel to AE, and we have the resultant in magnitude, 
direction, and position. 

, For join AC, AD. Then AC is the resultant of AB, 
BC, that is, of F l3 F 2 ; AD is the 
resultant of A C, CD, that is, of F 1 f 
F n , F 3 ; AEis the resultant of AD, 
DE, that is, of F X ,F % ,.P % ,F A \ which 
proves the proposition. 

We might have combined the two 
d diagrams as shown in Fig. 28, or we 

might have derived the resultant directly from the paral- 
lelogram of forces as shown in Fig. 29. 

Notice that in any method we may take the forces in any 
order, and we shall always find the same Fig. 29 

value of AE. Test this statement by 
making drawings to a large scale. 

Ex. 1. Three forces of 6, 8, 10 pounds 
act at angles of 120° with each other : 
find their resultant. Draw to scale by 
different methods, and compare results, 
compare. 

2. Forces of 1, 2, 3, 4, 5, 6 act at angles of 60 
Test as in Ex. 1. 

3. Forces of 20, 20, 21 pounds act at a point. The angle 
between the first and second is 120° and between the second 
and third 30°: find R. Ans. 29 pounds. 

4. Is it necessary that force and construction diagrams be 
drawn to the same scale ? 





Vary order, and 
findi?. 



50 DYNAMICS OF A PARTICLE. 

5. In the construction diagram the lines are drawn paral- 
lel to the forces : would it be allowable to draw them per- 
pendicular to the forces, or inclined at any (the same) angle, 
6 for example ? Test by a drawing. 

6. If the forces are in the same straight line, what does 
the force polygon become ? what the construction diagram ? 

49. Resolution of Forces. — By means of the parallelogram 

of forces, a force R can be found equivalent to two forces 

F t , F i} acting on a particle A. Conversely, 

c _' d t^g f orce R acting at A may be resolved 

/ Rs'i m *° two component forces F x , F 9 acting at 
7 ^S* J A, by constructing on R as diagonal a par- 

l/' { allelogram, and taking the sides to represent 

Fl E the components. The problem is similar to 
that already discussed in Art. 20. 

Ex. 1. If a force is resolved into two components, prove 
that the greater component always makes the smaller angle 
with the force. 

2. Resolve a force of 20 pounds into two components each 
of which makes an angle of 60° with it. 

Ans. Each = 20 pounds. 

3. Resolve a force of 10 pounds into two equal compo- 
nents, one of them making an angle of 45° with the force. 

Ans, 7 pounds, nearly. 

4. Find that rectangular component of a force of 10 
pounds which makes an angle of 60° with the force. 

Ans. 5 pounds. 

5. Explain the boatmen's saying, that there is greater 
"power" in hauling a canal-boat with a long rope than with 
a short one. 

The values of the rectangular components X, Y oi a force 
R may readily be computed analytically. Fi 31 

Thus in the right-angled triangles ABD, c '_ d 

A CD, ^/\ 

X=AB=R cos 6, " >^ 

Y= AC= 72 sin = R cos 90° - 6, \^e I 

which give the values of the two compo- A B 

nents X and Y. Hence the rectangular component of a 



RESOLUTION OF FOBCE3. 



51 



force R in a given direction is equal to R X cos (angle 
between component and R). As a check, 

X 2 + Y 2 = R 2 cos 2 6 + R 2 sin 2 6 = R 2 , 

which is also evident from the figure. 

Ex. 1. Find the components of a force 10 when 8 = 60°, 
90°, 120°, 180°, 240°, 270% 300°. Ans. 5,51^3; 0,10; 
— 5, 5 V3; — 10, ; - 5, — 5 Vs ; 0, — 10; 5, — 5 \/3. 
[Draw a figure for each case, and explain the sign of the 
result. ] 

2. The rectangular components of a force are each equal 
to p poundals : what is the force ? 

3. Show that the components of a force F in two direc- 
tions making angles f3, y with it are F sin yff/sin (/? -\- y), 
Fsin y/sm (ft -j- y), 

4. In a direct-acting steam-engine the piston pressure 
P is equivalent to P tan 6 perpendicular to its line of 
action and P sec 6 along the connecting-rod, being the 
angle of inclination of the connecting-rod to the line of 
action of the piston (see Fig. 119). 



Fig. 32 



50. The analytical solution leads us to a method of corn- 
lining forces which is often more conven- 
ient than the graphic method given in 
Art. 48. The two methods may be used 
to check one another. 

Take three forces F l3 F 93 F % , acting on 
a particle O. Through O draw any two 
lines OX, OY&t right angles to each other, 
and let l3 6 2 , 6 3 denote the angles which 
the directions of F l3 F 2 , F z make with OX. 
nents of 




Cos Q 



The compo- 



F x are F x cos 6 X along OX; F x sin 6 X along OT; 
F t are F 2 cos # 2 along OX; F 2 sin # 2 along OT; 
F 3 are F 3 cos 3 along OX; F 9 sin 0, along OY. 



52 DYNAMICS OF A PARTICLE. 

The components along OX being in the same straight line, 
may be combined by addition into a single force X (Art. 48) ; 
that is, 

F x cos 6 X + F t cos 6> 2 +^3 cos 6 9 = X. . . . (1) 

Similarly, the components along OY, being in the same 
straight line, may be combined into a single force Y, or 

F x sin 6 X + 2? sin 8 2 + ^, sin 0, = K . . . (2) 
Hence the original forces are equivalent to two forces X, Y 
acting in directions OX, OY at right angles to each other. 
The resultant of X, Y must therefore be the resultant of the 
original forces. Call it R, and let 6 be the angle it makes 
with the axis of X ; then 

R cos 6 = X, R sin 6 = Y. (3) 

Square and add (remembering that cos* 6 + sin 3 6 = 1), and 

R = VX* + Y\ 
which gives the magnitude of the resultant. Divide the 
second of equations (3) by the first, and 

tan 6 = Y/X, 
which gives the direction of the resultant. 

Hence, since the resultant acts at 0, it is known in posi- 
tion, magnitude, and direction, and is completely determ- 
ined. 

If we equate the values of X, Yin equations 1, 2, 3, we find 
RcosO = F x cos d x -f F 9 cos 2 + F 3 cos 3 ; 
R sin 6 = F x sin 6 X + F t sin 2 + F 9 sin 3 . 
Now OX, OY are any two rectangular axes. Hence the 
component in any direction of the resultant of a number of 
forces is equal to the sum of their components in the same 
direction. 

Ex. 1. Three forces of 6, 8, 10 pounds act on a particle at 
angles of 120° to each other: find the resultant in magni- 
tude and direction. 



RESOLUTION OF FORCES. 



53 



[Since the direction of OX is arbitrary, we may take it 
along any of the forces. Fig, 33 

(1) Take OX along the force 6. Then 
X = 6 - 8 cos 60° - 10 cos 60° = - 3; 
Y= 8 cos 30 ° - 10 cos 30° = - ¥% ; v 

and R = V9 +_3 = 2V3; , 

tan 6 =— 4^3/ _ 3 = 1/ ^3 or = 210°. 

(2) Take OX to fall along the force 8. Then /£ ' 
X = 8 - 6 cos 60° - 10 cos 80° ==_0 ; / 
Y= 10 sin "60° - 6 sin 60° = 2 V3; 

E = 2V3, as before; 
tan 6 = 2 V3/0 = 00 and = 90°; 
or the resultant is perpendicular to the force 8, showing it 
to be in the same relative position with reference to the 
other forces as before.] 

2. Solve with OX along the force 10. 

51. Having now the means of combining the forces that 

act on a particle into a single force giving the same motion, 

Fig. 34 y "we proceed to study this motion. 

ty* r^_^_ Conceive a particle O acted on by 

-&^~- — c^r^CIII^ a number of forces whose result- 

^^ ant is R. If a force equal and 

opposite to R be added, the whole system of forces acting 

on the particle will balance. The resultant of the forces is 

nil, and the system is said to be in Equilibrium. From the 

relation F= ma, it follows when F= that a = 0. Hence 

a system of forces in equilibrium implies that there is no 

acceleration. Thus the velocity of the body, if it had any 

before the forces commenced to act, would be unchanged, 

and the motion would continue uniform and in a straight 

line; if at rest, it would remain at rest. Equilibrium, 

therefore, does not imply rest, but rest implies equilibrium. 

The branch of dynamics which considers the circumstances 

for which equilibrium is possible is called Statics. 

When the forces do not balance, an acceleration arises 
from the resultant force, and the particle moves with a 



54 DYNAMICS OF A PARTICLE. 

motion compounded of the motion in its original path, and 
that due to the resultant. The branch of dynamics which 
considers the circumstances under which change of motion 
takes place is called Kinetics. 

STATICS OF A PARTICLE. 

52. When a particle is in equilibrium under forces acting 
in the same straight line, the total acceleration produced 
by the forces is nil, and therefore the sum of the acceler- 
ations in one direction is equal to the sum in the opposite 
direction. Hence the sum of the forces acting in one di- 
rection mast be equal to the sum in the opposite direction. 
In other words, the forces must reduce to two forces equal 
in magnitude and opposite in direction. 

Next, let three forces not in the same straight line act on 

c the particle. Find the 

Fig. 35 r/\ resultant R of any two 

F a/^^R \ Fi>F*° For equilibrium 

^ A it ;- >D to exist, R and F 9 must 

/JN. y 3 

jK y^ De equal and opposite. 

1 ]^ Hence if three forces 

c acting on a particle are 

%^$\\ f in equilibrium, any one 
s' -i \. of them is equal and op- 
a f ' d posite to the resultant 

of the other two. 
The sides of the construction triaugle A CD are parallel 
to the three forces F j9 F a , F 2 , and proportional to them 
in magnitude (Art. 47). Notice that their directions are 
the same way round the triangle. Hence three forces act- 
ing on a particle ivill be in equilibrium if they can be rep- 
resented by the three sides of a triangle drawn parallel to 
the forces, and taken the same way round. This proposi- 
tion is known as the Triangle of Forces. 

It may be expressed analytically. If a, /?, y be the 



POLYGON OF FOEOES. 



55 



angles between the directions of the forces, then in the 
construction triangle the angles are evidently 180° — a, 
180° — /?, 180° — y\ and since the sides of a triangle are 
as the sides of the opposite angles, 

DC/am (180° - a) = CA/sm (180° - fi) 

= AD/$m (180° - y) 9 
or 

jP,/sin a = FJsin j3 = F 3 /sm y\ 

that is, when three forces acting on a particle are in equi- 
librium, each is proportional to the sine of the angle between 
the directions of the other two forces. 

Illustration. — Take a piece of board, and drive in three 
smooth pegs A, B, C, or place 
three pulleys at A, B, C. Run 
strings over the pegs, and knot 
together as at O. Suspend 
weights from the strings. 
Draw lines along the strings on 
the board, and plot the triangle Hn 
ale with sides parallel to these 
lines. The sides of this tri- 
angle will be found to be pro- 
portional to the weights. 

Ex. 1. Make the weights 3, 4, 5 oz., and it will be found 
that one angle of the triangle abc will be 90°. 

2. Could the three weights be equal to one another? 
Plot abc in this case, if possible. 

53. Polygon of Forces. — If in the force diagram of Art. 
48 the direction of R be reversed, the particle will be in 
equilibrium under the action of F l , F 2 , F s , F 4 , — B. In- 
dicate the directions of these forces on the construction 
diagram, and notice that they are the same way round. 
-Hence any number of forces in the same plane acting on a 
particle ivill be in equilibrium if they can be represented by 




56 DYNAMICS OF A PARTICLE. 

the sides of a polygon drawn parallel to the forces, and 
taken the same way round. This is known as the Polygon 
of Forces. 

Ex. In the polygon of forces any side represents in mag- 
nitude and direction the resultant of the remaining forces, 
but with sign reversed. 

54. The analytical equivalent of the polygon of forces 
may be deduced from Art. 50. For if the forces acting at 
are in equilibrium, the resultant R must be equal to 
zero. Hence 

x* + r 2 = o, 

which, since X 1 and F 2 are both positive, can only be satis- 
fied by X = 0, Y = 0, that is, by 

F 1 cos 0, + F % cos 2 + . . . = 0, 
F l sin 0, + F 9 sin a + . . . = 0. 
Hence if any number of forces in the same plane acting 
on a particle are in equilibrium, the sums of the compo- 
nents of the forces along any two straight lines at right 
angles to each other through the particle are equal to zero. 
Ex. 1. State the analytical conditions of equilibrium, 
i when 2, 3, ... 7i forces act on a par- 
ticle. 

2. A rod AB whose weight may be 
neglected is hinged at A, and supports 
a weight W at B. It is held up by a 
string BC fastened to a fixed point C 
vertically above A. If AB is horizontal 
and angle ABC — 30°, find the tension 
T x of the string, and the thrust T 2 along 
AB. 

[The point B is in equilibrium under T l9 T^ y W. Re- 
solve vertically and horizontally, then 

T x cos 60° - W = 0, 

T x cos 30° - 7 T 2 = 0; 

.\ T 1 = 2W i 1\= W Vz.) 

3. In a canal with parallel banks, a boat is moored by two 

ropes attached to posts on the banks. If the ropes are in- 




KIXETICS OF A PARTICLE. 57 



dined at angles of 30°, 60° to the banks, compare the pulls 
on them, both ropes being in the same horizontal plane. 

Ans.l'.Vs. 



KINETICS OE A PARTICLE. 

55. If a number of forces act on a particle and the re- 
sultant be found, a certain motion is due to this resultant. 
If the particle has this motion, it is said to he free; if it has 
some other motion, the deviation must be owing to the 
entrance of some cause not accounted for, and the motion 
is said to be constrained. In free motion the particle is 
isolated from all causes tending to affect its motion except 
the acting forces, while in constrained motion this is not 
the case. 

We have seen (Art. 5) that the position of a particle is 
defined by its coordinates with reference to certain axes 
assumed to be fixed. A change in position is represented 
by changes in these coordinates. Hence the coordinates 
being either a distance and two angles or three distances, 
a point is said to have three degrees of freedom to move. 

If the point is compelled to remain in an assigned plane 
(as the plane of the paper), its position is defined by two 
coordinates, and it is said to have two degrees of freedom 
and one degree of constraint. Similarly, if compelled to 
remain at the same distance from a fixed point it would 
move on the surface of a sphere, and have two degrees of 
freedom and one of constraint. 

Again, if the point were compelled to remain in two 
planes, that is, in their line of intersection, it would have 
one degree of freedom and two of constraint: so also if 
compelled to remain in one plane and keep at the same 
distance from a fixed point, that is, to move in a circular 
path. 

If compelled to remain in three planes, it can have only 



58 DYNAMICS OF A PARTICLE. 

one position, their point of intersection, and is therefore 
wholly constrained. 

Ex. How many degrees of freedom has a curling-stone 
on smooth ice; a stone in a sling; a compass, joined to a 
tripod by a ball-and-socket joint ? 

56. Free Motion. — By means of the relation F = ma 
connecting force acting, mass acted on, and acceleration 
produced, we are able to extend the geometrical properties 
of motion to particles acted on by given forces. Various 
paths may result, depending on the motion of the particle 
at the time the force begins to act. We shall first of all 
consider the particle to be unconstrained, and have all de- 
grees of freedom. 

Suppose the particle to have an initial velocity u, and 
that the force i^acts in the direction of this velocity, caus- 
ing an acceleration a. Then the resultant velocity at the 
end of a time t is composed of that due to u and that due to 
the acceleration a. Hence if v 'is the final velocity and s 
the distance passed over, we have (Art. 13) 

v = u -}- at, s = id -f- \at*, 

= u + Ft/m, = ut + iFF/m, 

in terms of the absolute units. 

Ex. 1. A mass of 10 lbs. is moved along a smooth table 
by a weight of 6 pounds attached to a string which passes 
over a smooth peg on the edge of the table: find the dis- 
tance passed over in 2 sec, and the velocity acquired. 

[Effective force = G pounds == 6(/ poundals ; mass moved 
= 10 + 6 = 16 lbs.; .-. a = 6g/16 = 12 ft.; v = 2 xV2 = 
24 ft. per sec; s = J X 12 X 2 2 = 24 ft.]. 

2. An ice-boat weighing 1000 lbs. is driven for 30 sec 
by a force of 100 pounds: find the velocity acquired and 
the distance passed over, supposing it starts (1) with a ve- 
locity of 10 ft. per sec, (2) from rest. 

3. Find the tension P of the string in Ex. 1. 

[Effective force on mass 10 = 10 q — P; , 

.'. accel. = (Wg - P)/10. 



FREE MOTION. 59 

Effective force on mass 6 = P — 6g; 

.: accel. = (P -6g)/6. 
Hence P = 7-| pounds.] 

4. An elevator weighing m lbs. is lifted by a force of n 
pounds: find the acceleration and tension of the lifting 

, '. . , m — n „ 2mn 

chain. Ans. a = ; a; P = : — q. 

m-\- n J m -\- n 

5. A bucket weighing 25 lbs. is let down into a well with 
a uniform velocity: find tension of rope. 

Ans. 25 pounds. 

6. Two bodies of weights ft^and a\ pounds are Fig. 38 

fastened to a string which passes over a smooth 

peg: find the acceleration and tension of the 

' . . w 1 — w s 2 1,1 
string. Ans. a = - i — ■ -a ; — - = — -f — . 

[If the value of a is observed, we have the 

value of a from — 1 — . all of the quanti- 

* w x — w ? ^ 

ties in this expression being now known. By 
making the differences between w x and tv 2 small, 
the acceleration a may be made as small as we 
please. The smaller it is the easier it is to ob- 
serve. In this consists the advantage of using 
two weights instead of a single weight falling 
freely to determine g. 

To find a we observe the distance s passed over in t sec. 
by means of a graduated scale placed vertically. Then, 
since s = ^at 2 , we have a at once. 

For example, let ic\ = 21 oz., to 2 = 20 oz., and suppose 
that in 5 sec. the weight w has fallen 9. 8 ft. Then 

19 6 004.01 

a = -^- = 0. 781, and g = ^ ^ ; X 0.781 = 32 + ft. 

An apparatus constructed on this principle, but with 
certain mechanical contrivances for lessening friction and 
giving convenient means of measuring the heights fallen 
through, is known as At wood's machine. It is to be found 
in most physical laboratories.] 

7. A train of 100 tons is running at the rate of 15 miles 




60 DYNAMICS OF A PARTICLE. 

an hour: find what constant force is required to bring it to 
rest in (1) one minute, (2) one mile. 

Ans. (1) 220,000 poundals ; (2) 165,000 poundals. 

8. What pressure will a man weighing 150 pounds exert 
on the floor of an elevator descending with an acceleration 
of 4 ft. per sec. ? Ans. 131i pounds. 

9. How is it that as an elevator comes to rest in its de- 
scent one feels as if he were being lifted up ? 

10. A man who is just strong enough to lift 150 lbs. can 
lift a barrel of flour of 200 lbs. from the floor of an elevator 
while going down with an acceleration of 8 ft. per sec. 

11. The pull of the engine on a train whose weight is 100 
tons is 1000 pounds. In what time will the train acquire 
a velocity of 45 m. an hour? Ans. 7 min. 42 sec. 

12. An express engine weighing a lbs. starts from a depot 
with a train of n cars of b lbs. each, with an acceleration 
of c ft. per sec: find (1) the pull the engine is exerting; (2) 
the pulls on the successive couplers from the engine to the 
rear of the train. 

Ans. Pull on engine coupler = ten poundals. 

57. Falling Bodies. — A case of special interest is when the 
acting force is the force of gravity. This force acts verti- 
cally downwards, and being constant, produces a constant 
acceleration g ( = 32.2 ft. or 981 cm.) vertically downwards. 
Hence if a particle has a velocity u, its motion is com- 
pounded of two motions, — one due to this velocity, and the 
other to the acceleration g. Suppose the velocity u to be 
vertical. Then the two motions are in the same vertical, 
and we have the case of falling bodies. The resultant ve- 
locity v, at the end of a time t, would be the sum of the 
original (or initial) velocity u and the velocity gt acquired 
under the force of gravity in this time, that is, 

v = u ± gt, 

according as the initial velocity of projection u is vertically 
downwards or vertically upwards. 



FREE MOTION". 61 

Also, as in Art. 13, we have 

s = ut ± ye 

for the distance fallen or height acquired in the time t. 
By eliminating t we have a direct relation between v, g, 

v* = if ± 2gs, 
which is often convenient. 

If the body falls from rest, there is no initial velocity, and 

v = gt, s = ± gt\ v 2 = 2gs. 

This naturally follows from the statement in Art. 36, 
that the acceleration contributed by the force of gravita- 
tion is independent of the mass of the body.* Hence the 
whole question is one of kinematics. 

It is of course understood that the action of the atmos- 
phere is not taken into account. The motion is conceived 
to take place in a vacuum. 

58. The general equation of motion for a falling body 
would be 

tfs/df= -g, 

which may be developed as in Art. 14, and the results found 
above will be obtained. 

Ex. 1. A body is projected vertically upwards with a 
velocity of 161 ft. per sec. : find (1) when it will come to 
rest; (2) the height to which it will rise. 

[(1) When it comes to rest v = 0; . *. substitute v = 0, 
u = 161, g = 32.2 in v = u — gt and t = 5 sec; (2) sub- 
stitute t — 5, u = 161, in s = itt — igt* and s = 402.5 ft. 
Or substitute v = 0, u = 161 in v 2 = if — 2gs and s = 
402.5 ft.] 

* Galileo (1564-1642) was the first who appealed to experiment in 
all physical inquiries. Until his time it was taught that the velocities 
of falling bodies are proportional to their weights. He argued that 
if this were true two crown pieces must fall faster when sticking to- 
gether than when unconnected, which is contrary to experience. 



62 DYNAMICS OF A PARTICLE. 

2. In Ex. 1, find the velocity when the body is at a height 
of 257.6 ft. 

Ans. v — ± 96.6 ft. per sec. ; the plus sign indicating the 
velocity of ascent, and the minus sign the velocity of de- 
scent, each at 257. 6 ft. from the point of projection. 

3. A body is projected vertically upwards with a velocity 
of 161 ft. per. sec; find at what times it is 257.6 ft. above 
the starting point. Explain both answers. 

4. In Ex. 3, find the total time of flight. 

5. Find the distance passed over by a body falling freely 
during the sixth second of its fall. Ans. 176 ft. 

6. It is required to project a body vertically to a height 
of 36 ft. ; find the velocity of projection. Ans. 48 ft. per sec. 

7. A stone thrown vertically upwards is observed to be at 
a height of 96 ft. in 2 sec; how much higher will it rise? 

Ans. 4 ft. 

8. Two bodies are dropped from a height at an interval 
of 2 sec ; find the distance between them at the end of the 
next 2 sec Ans, 192 ft. 

9. A falling body describes sit. in the nth. sec. of its fall; 
show that the initial velocity is s — g(n — 0.5) ft. per sec. 

10. If s l9 s i9 s 3 are the distances described by a falling 
body in t l9 t 29 t 2 seconds, prove that 

S l (t,-t,)+8,(t,-t l )+8,(t,-t,) = 0. 

59. Projectiles. If a particle is ^projected from a given 
point in a given direction, and is acted on by the force of 
gravity only, it is called a projectile. If the direction of 
projection is vertical, the initial velocity and the accelera- 
tion due to gravity are in the same direction and the path 
is a straight line. This case has already been considered 
under falling bodies. 

But if the direction of projection is not vertical, the path 
is a curve which must be in the vertical plane containing 
the direction of projection, there being no force to cause it 
to move out of this plane. This case we proceed to discuss. 

Let the velocity of projection u make an angle 6 with the 
horizontal line OX. Resolve u into vertical and horizontal 



FREE MOTION. 



63 



components, that is, into u cos 6 along OX and u sin 6 
along OY. The total vertical velocity v, , at the end of a 



Y 

OS 

1 


1 

85 




Fig. 39 

c 


7 


i 


p u cos e 






uCosd 


B AX 



time t when the particle has reached a point P, say, is 
(Art. 57) 

»,= z< sin 6 — gt, (1) 

and the distance PQ( = ij) passed over in a vertical direc- 
tion is 

y = tic sin 6 — J gf (2) 

The horizontal velocity r, is, since gravity has no effect on 
the horizontal motion, 



v^= ucos 6, 



(3) 



and the distance OQ (=x) passed over at the end of the 
time t is 

x = tu cos 6 (4) 

From these four equations the motion is determined. 

To find the path of the particle. By ascribing to t differ- 
ent values and computing the values of x and y the positions 
of as many points in the path can be found as desired. Or 
by eliminating t between equations (2) and (-A) we find a 
relation between the co-ordinates x, ij which holds for all 



64 DYNAMICS OF A PARTICLE. 

values of t and is therefore the equation of the path. This 
gives 

y — x tan 6 — ga?/2u* cos 2 6, 

which represents a parabola.* 

To find the time of flight, that is, the time in which the 
particle will reach the line OX. When this happens y = 0, 
and 

.-. = tic sin 6 - \gt\ 

whence t — 0, t = 2u sin 6/g; 

which shows that the particle is twice on the line OX, once 
at 0, the beginning of the motion, when t = 0, and again at 
A, the end of the motion, when t = 2u sin 6/g. The hitter, 
being the time from the beginning to the end of the motion, 
is the time of flight. 

The horizontal distance OA, or the range, is the value of 
x at the end of the time of flight. Hence 

range OA = u cos 6 X 2u sin 6/g = u* sin 2 6/g. 

The greatest value sin 26 can have is unity, and this oc- 
curs when 26 — 90° or 6 — 45°. Hence the horizontal 
range is greatest when the angle of projection is 45°. This 
result is not true in practice, as we have not taken into 
account the resistance of the air. Experiment gives an 
angle of about 34° instead of 45°. 

To find the greatest lieiglit consider that at the highest 
point the vertical velocity is nil. For, if not, the particle 
could rise higher. This gives the relation 

= n sin 6 — gt 
and 

t = u sin 6/g, 

* This is the great discovery of Galileo. No attempt had beeu 
made up to his time to explain curvilinear motion of any kind. 



FREE MOTION. 65 

showing that the greatest height is reached in half the time 
of flight. Substitute in the value of y which gives the 
height at any time this value of t, and 

greatest height BO = u* sin 2 0/2g. 

Also, OB = u cos 6 X u sin 6/g = u 2 sin 2 6/2g, 

which is half the range. 

The resultant velocity v at any point P on the path is 



v = Vv* + v* 



= Vu 2 -2ugtsmO + gH 2 ' 

It will be noticed that, as in the case of a falling body, all 
results are indepeudent of the mass of the body. Hence 
they are true whether the mass projected is large or small, 
and the problem may be considered as kinematical. 

60. The differential equations of motion in the case of a 
projectile are 

eTx/d? = 0, cVy/df = - g, 

x, y being the coordinates of the particle in its path at the 
end of a time /. Integrating between the limits and t, 
remembering that v cos 6 is the initial velocity along the 
axis of X, and v sin 6 that along the axis of Y we have 

dx/dt = v cos 6, dy /dt = v sin 6 — gt. 

Integrating a second time, 

x = tv cos 6, y = -fv sin Q — \ gt 2 , 

the results already found. 

Ex. 1. If the velocity and acceleration take place in the 
same straight line, what is the equation of the path ? 

2. A balloon floating in a horizontal current of air with 



66 DYNAMICS OF A PARTICLE. 

a uniform velocity of 45 miles per hour suddenly collapses 
and descends with an acceleration of 32 ft. per sec. ; trace 
its path. 

3. A ball is thrown horizontally from a height of 10 me- 
ters with a velocity of 20 meters per sec: find when it strikes 
the ground; the range; the final velocity; and the inclina- 
tion of the direction of motion at the point of striking the 
ground. 

4. Prove that the range for an elevation of 30° is the 
same as for an elevation of G0°, and for an elevation of 
45° -f- 6 the same as for 45°— 6. 

5. Compare the greatest elevations in the two cases. 

6. Prove that the height of the vertex in feet is nearly 
4 times the square of the time of flight in seconds. 

7. A body is projected horizontally from a given height 
li with a velocity u\ prove that the equation to the path is 

2u*y = gx*. 

Show that the range is equal to u Vh/i nearly. 

8. A ball is fired at an angle of 45° so as just to pass over a 
wall 10 ft. high at a distance of 100 yards. How far from 
the wall will it strike the ground? Ans. 10.34 ft. 

9. "Swift of foot was Hiawatha: 

He could shoot an arrow from him 

And run forward with such fleetness 

That the arrow fell behind him ! 

Strong of arm was Hiawatha : 

He could shoot ten arrows upward, 

Shoot them with such strength and swiftness 

That the tenth had left the bowstring 

Ere the first to earth had fallen." 
If one second elapsed between the discharge of each of the 
arrows and Hiawatha shot at his greatest range, prove that 
he must have been able to run at the rate of 99 miles an 
hour. 

10. Show that for parabolic motion the hodograph is a 
straight line. 

61. Central Forces. — If the acting force .Pis constantly 
directed towards a fixed point or centre it is said to be central. 



FREE MOTIOX. 



67 



Fig. 40 



The most important case is that of attractive forces in which 
the law of attraction is that of the in- 
verse square of the distance r of the par- 
ticle from the centre, or F = C/r 2 where 
C is a constant. 

Let be the centre; OX, OY the 
axes of coordinates; and x, y the coor- 
dinates of the particle F at a time t. 
Let OP = r, m the mass of the particle, and the angle 

pox- e. 

The general equations of motion are 



cFx 
df ' 



F 



cos 6, 



d 2 y F . „ 

-172= sin 6, 

dt m 



d 2 x _ ex . . 

df - ~ P 7 ' • ' " {) 



d ll - _ c l 
df r* : 



(3) 



if we place F/m = C/r 2 m = c/r 2 where c = C/m. 

The relation between x and y will give the equation to 
the path of the particle. To find it: 

Multiply the first equation by y, the second by x, and 
subtract. 



X dt 2 ' 



y 



cVx 
df 



and by integration 



.fy 



dx 

x tt-y-df 



0; 



h, a constant. 



Also, since x = r cos 0, y = r sin 6, 



^ dy dx 



dt 



y 



dt ' 



,dd 
r df 



(3) 



68 





DYNAMICS 


OF 


A PARTICLE. 








e, eliminating r a , 














d\c 
df ~ ' 


c a dd 




d*y 
df ~ 


c 
~k 


sin 


6 


dd 
~di 



and by integration 



dx c a .^ q dy 

di 



c . a dy C 

77 — c, = — 7 sin 0, -± — c= T cos 6 
k dt a k 



= - cy/kr, (4) = cx/kr, (5) 

when c x , c 2 are constants. 

Multiply the fourth equation by y, the fifth by x, and sub- 
tract, and we have 

k -\- c x y — c 2 x = cr/k = c Vx 2 -f- y*/k, 

the equation to a conic section with the origin at the focus. 
Hence the path described by a particle under the action 
of a central force varying inversely as the square of the 
distance is a conic section, whose focus is the centre of 
force. 

. This is the case of planetary motion, the sun being at 
the centre of force. 

The further discussion of this problem will be found in 
works on mathematical astronomy.* 

62. Constrained Motion. — To a particle acted on by a 
force F in an assigned direction a certain path results. If 
the path differs from this, it must be owing to some cause 
which changes the motion, that is, to the action of another 
force. Hence if the path is prescribed we may, by adding 
forces which with the original force will give a resultant 
which can produce this path, consider the motion free. 
The discussion will therefore come under the principles 
already laid down. 

* The development of this subject is due to Sir Isaac Newtou. 



COKSTKAIXED MOTION. 



69 



•Resolve every given 

Fig. 41 



F Sin 6 



63. Motion on a Horizontal Plane, 
force F into its two compo- 
nents F cos 8 along the plane 
AB and F sin 8 at right 
angles to it, 8 being the in- 
clination of F to the plane. 
To each of these forces an 
acceleration is cine. Bnt the particle is constrained so as 
not to move in the direction of the force i^sin 8. This can 
be brought about by assuming that the plane exerts an 
equal force F sin 8 in the opposite direction. 

As regards the horizontal stress between the particle and 
the plane, we can say nothing a priori. Experiment shows 
that it depends on the nature of the surfaces in contact. 
We shall for the present assume that the stress between the 
particle and the plane is normal only, or, as it is often ex- 
pressed, that the plane is smooth. 

If therefore a particle slides on a smooth plane under 
the action of a force F inclined at an angle 8, the reaction 
of the plane is F sin 8 and the force acting along the plane 
is F cos 8, which latter being that to which the motion is 
due, is the effective force. 

64. Motion on an Inclined Plane. — Suppose a particle of 
Fig. 42 ^b mass m lbs. on an inclined plane, 

and acted on by gravity only. 
The force acting is mg poundals 
vertically downward. 

Resolve mg into components 
mg sin 8 along the plane, and 
mg cos 8 at right angles to it. The 
reaction of the plane is mg cos 8, 
and the effective force along the plane is mg sin 8. The ac- 
celeration along the plane is therefore mg sin 8/ni — g sin 8. 
If the initial velocity is in the direction of the accelera- 




70 DYNAMICS OF A PARTICLE. 

tion along the plane, the path is a straight line. If u de- 
note the velocity at B, the velocity v attained on reaching 
A at the foot of the plane is given by 

v* = if -f 2g sin 6 x AB 

= it 2 -\- 2gh, 

when li is the height of the plane. Hence the final veloc- 
ity is independent of the inclination of the plane. 

If the initial velocity is not in the direction of the acceler- 
ation along the plane, the path is a parabola, whose equation 
may be found from Art. 59 by substituting g sin 6 for g. 

65. The above results may also be deduced from the 
general equation of motion of a particle on an inclined 
plane, 

d's/df = -g sin 6, 

where is the inclination of the plane to the horizontal, 
and s the distance of the particle P from the point at the 
time t. The solution is similar to that of Art. 14. 

Ex. 1. A body starts from rest and falls down a plane of 
height h: prove that the velocity acquired is V'2gh and the 
distance passed over in / sec. is \gf sin 6. 

2. Prove that the velocity of a particle on reaching A 
(Fig. 42) from B by moving on the plane is equal to that 
acquired by falling freely through the height h of the 
plane. If t 1 , t 9 are the times required to attain these veloci- 
ties, prove t 2 — t 1 sin 0. 

Pi g# 43 3. Prove that the times of descent of a 

particle starting from the extremity A of 
a vertical diameter AL is the same along 
all chords A B, A C, . . . of the circle. 

Hence find the line of quickest descent 
from a given point to a given straight 
line. 

4. Find the line of quickest descent 
from a given point to a given vertical 
circle. 




Fig. 44 



CONSTRAINED MOTION. 71 

66. In the case of a particle of mass m sliding down a 
plane from rest, the motion along the plane is uniformly 
accelerated, the moving force being 
mg sin 6. For equilibrium a force 
must be applied such that the resultant 
force along the plane is nil. Hence 
if a force F be applied parallel to the 
plane, it will hold the particle in equi- 
librium if 

F = mg sin 9 or F : mg = BO:AO = height : length.* 

If the force F be applied 




Fig. 45 




parallel to the base, the con- 
dition of equilibrium would be 
F cos 6 = mg sin 6, 
or F'.mg = BC'.AB 

= height : base. 
The normal pressure in the 
first case would be 



JV = mg cos 6, 
and in the second 

]Sf = mg cos -f- F sin 0~— mg sec 6. 

Ex. 1. Show that the values of F and JVare the same if 
the particle moves uniformly upward or downward, or is at 
rest. 

2. Show that the force F is most efficient when acting 
parallel to the plane. 

3. A weight of m lbs. is placed on a smooth inclined 
plane, and is acted on by a horizontal force of mp poundals : 
find the acceleration. Ans. a = (g sin ± p cos 6). 

4. On an inclined plane a horizontal force F supports a 



* This is of special interest, ns being the problem of oblique forces 
first solved. The solution is due to Simon Stevinus of Bruges, Bel- 
gium (1548-1620). It may be found in Wheioell, Mechanics, p. 44. 



72 DYNAMICS OF A PARTICLE. 

weight W, and a force Q parallel to the plane will also sup- 
port W; prove 

5. On an inclined plane a force P acting parallel to the 
plane can support a weight W x , and acting horizontally a 
weight W a ; prove 

117 - W* = P\ 

67. Motion in a Circle. — Suppose a particle of mass m 
to move with constant velocity in the circumference of a 
circle of radius r. As in Art. 23, draw the hodograph of 
the particle, and let this be the circle aba with centre 
(Fig. 16). The velocity at a is perpendicular to On which 
is parallel OA and equal in value to a constant quantity a. 
But the velocity at any point of the hodograph is equal to 
the acceleration at the corresponding point of the original 
path. Hence the acceleration in the path ABC is always 
directed to the centre of the path; in other words, the 
particle as it moves in its circular path is acted on by a 
constant force F directed to the centre of the circle, and 
therefore with its direction always perpendicular to the 
direction of motion. 

To find the magnitude of this force. Let t be the time 
in which the circle ABC is described, then tv — Inr. Also, 
since the circle abc is described in the same time, ta — 2nv. 
Hence by eliminating t 

a = v 1 /r 
and F— ma — mv*/r. 

To this force F acting constantly towards the centre of 
the circle as the particle moves uniformly in the circumfer- 
ence the name of Centripetal Force is given. The velocity 
being constant, the acceleration contributed by the centrip- 
etal force is completely spent in changing the direction of 
motion from point to point. 



COXSTRAIXED MOTIOX. 73 

It may be illustrated by supposing a particle attached by 
a string to an axis through and made to revolve about 
this axis with uniform velocity in a region from which the 
effect of gravity is conceived to be eliminated. The circu- 
lar motion is thus due to the initial velocity v and to the 
pull of the string, and to these alone. 

Suppose that the string is cut when the particle reaches 
A. Since there is no force now acting, the particle will 
move in the straight line AA 1 tangent to the circle at A 
and with velocity v. This follows from the law of inertia. 
The constant pull of the string therefore acts only in 
changing the motion from uniform rectilinear to uniform 
circular motion. 

Suppose next the string removed and the constant pull 
replaced by a force F. This force must act towards the 
centre 0, must act constantly, must continually change its 
direction as the particle changes place, and must produce a 
constant acceleration a, which, combining with the constant 
velocity v along the tangent, changes at every moment the 
direction of the velocity v without changing its magnitude. 

Consider further. The pull in the string is a stress, the 
action on the particle towards the centre and the reaction 
of the particle from the centre. The two are equal. The 
first. is the centripetal force, and to the other the name Cen- 
trifugal Eorce* is given. The centripetal force being the 
force required for producing the change of direction, the 
centrifugal force is therefore really the inertia-resistance of 
the particle. 

68. The centrifugal force may be expressed in terms of 
the number of revolutions made in a given time. Thus if 
n is the number of revolutions per second, then v = 27tr?i 
and 

F = m x 47r 9 rV/r 

= 39.48 mm* poundals. 
* Term introduced by Huygens (1029-1095). 



74: DYNAMICS OF A PARTICLE. 

If n is the number of revolutions per minute, and we call 
the weight of the body w lbs., this reduces to 

F= 0.00034 wrn* pounds, 

the rule used in practice. ' 

69. A very remarkable application of the idea of centrip- 
etal force was made by Newton to test the truth of the law 
_. „ that the acceleration of gravity 

varies inversely as the square of the 
distance from the earth's centre. 
Observation shows that the moon 
revolves round the earth in an 
orbit nearly circular and with uni- 
form velocity. If v = velocity of 
moon, R = radius of orbit, then 
the acceleration of the moon di- 
rected to the centre of the earth 
is v*/R. If this acceleration is due to gravity, we have 

g' = v'/R 

when (f is the value of g at the distance R from the earth's 
centre. Also, if the acceleration of gravity varies inversely 
as the square of the distance from the earth's centre, 

g'/g = S/B> 

when r is the earth's radius. 

Hence, eliminating #', we have, as the condition to be sat- 
isfied if the hypothesis is true, 

v 2 R = r 2 g. 

Now from measurement, R = 240,000 miles, v = JOOO 
miles, g = 32 ft., time of revolution of moon = 27 days 8 
hours = 2,360,000 seconds nearly, whence v = 3375 ft. per 
sec. Substituting these values, the expression will be found 
to check. 




CONSTRAINED MOTION. 75 

Also, we may find the difference between the true and the 
apparent weight of a body at the equator as affected by cen- 
trifugal force. For the true weight being the pull of the 
earth if at rest differs from the apparent weight or actual 
pull of the earth by the centrif ugal force. Hence if m is 
the mass of the body, 

mg — mg 1 = mv 2 /r 
when g x is the actual acceleration due to gravity. If, for 
example, m = 1 lb., a simple reduction will show that the 
centrifugal force will amount to 24 grains, giving the dif- 
ference of weight. 

Also, we have 

vt = 27tr 
when t = 24 hours and r = 4000 miles. Hence 

g-g x = 0.111 ft., 
or the acceleration of a particle falling freely at the equator 
is 1/9 ft. less than it would be if the earth did not revolve 
on its axis. 

The actual acceleration of gravity at the equator is ob- 
served to be 32.09 ft. per sec. Hence if the earth did not 
revolve on its axis the acceleration of gravity there would 
be 32.09 + 0.11 = 32,20 ft. per sec. 

Ex. 1. Why cannot the centripetal force increase the ve- 
locity of the moving particle ? [• . • always perpendicular to 
the direction of motion.] 

2. Explain how it is that although the particle con- 
stantly gains velocity along the radius it never possesses 
any such velocity. 

3. A stone of If lbs. is whirled 90 times a minute at the 
end of a string 3^ ft. long: find the tension of the string. 

Ans. 544 poundals, or 17 pounds nearly. 

4. The distance of the moon from the earth is 240,000 
miles and she revolves round the earth once in 27 days S 
hours. Find the acceleration relative to the earth. 

5. Find the velocity of projection in order that a bullet 



76 DYNAMICS OF A PARTICLE. 

shot horizontally may travel round the earth continually. 
[From v'/r = g, v = 5 miles per second nearly.] 

6. Show that the centrifugal pressure on the rails by a 
locomotive of w lbs. moving at the rate of v miles per hour 
in a curve of r feet radius is 0.0GG9 wv*/r pounds. 

Qa. The driving-wheels of a locomotive are 5 ft. in diam- 
eter and the cylinders 2 ft. stroke. If the velocity is 50 
miles an hour, find the centrifugal force developed by the 
counterweight, supposing it to weigh 300 lbs. and that the 
distance of its centre of gravity from the center of the axle 
is 20 in. Ans. 13,328 pounds. 

8. Show that in lat. 60° the normal component of the 
centrifugal force of the earth's rotation is one fourth of 
what it is at the equator. 

0. Show that the centrifugal force at the equator is 
1/289 of what the force of gravity would be if the earth did 
not revolve on its axis. 

10. Show that if the earth were to revolve on its axis 17 
times faster than it does, all bodies at the equator would be 
without weight. 

11. Show that " a body weighing 1 lb. avoir, on a spring- 
balance at the earth's equator would weigh only 2.6534 
ounces upon the same spring-balance at the moon's equa- 
tor." 

70. Simple Harmonic Motion. — Suppose that while the 
particle P moves in a circle of radius r with uniform veloc- 
ity v, another particle M moves 
k ' 9 " along the diameter A B in such 

a way that PM is always per- 
pendicular to AB. Both P and 
M must start from the same 
point A, both will reach B at the 
same time, and while P makes 
a complete revolution, M will 
move from A to B and back 
again. The motion of M is called simple harmonic motion* 

* " Physically, the interest of such motions consists in the fact of 
their being approximately those of the simplest vibrations of sound- 




CONSTRAINED MOTION. 77 

(S. H. M.). The time of oscillating from A to B and back 
again, being the same as that required for passing once 
round the circumference in the corresponding circular 
motion, is 2nr/v, and is called the Period of the S. H. M. 

The velocity of M being equal to the component of the 
velocity of P in the direction ^4^ is v sin POA. The 
acceleration of M is the component of the acceleration 
of P in the direction AB. But the acceleration of P is 
along PO, and is equal to v*/r. Hence the acceleration 

of M=- cos POA •:= - a ; X OM, and therefore v 2 /r = 
r r 

acceleration of M/ M. The distance OM of the particle 

M from the centre is called the Displacement. 

We may therefore write 

period T ' = %7tr/v 

= 27T '/displacement/acceleration. 

71. This may e proved otherwise. Let the particle start 
from A, and let P be its position in the circular path at the 
end of a time t. The angle POA expressed in circular 
measure is 2fft/ T = cot, where go is the circular measure of 
the angle described in one second. Hence 

x = OM = OP cos POA = r cos cot. 
Also. v = dx/dt = — gov sin cot, 

and a = d 2 x/df = — otfr cos cot = — gd^x, 

or the acceleration along OX varies as the displacement x, 
the result found above. 

72. The distance OA or OB of the extreme position of 
the particle from the mean position is called the Ampli- 
tude of the S. H. M., and the fraction of the period which 
has elapsed since the particle M left its initial position A is 
called the Phase of M at the time t. 

ing bodies, such as a tuning-fork or pianoforte wire, — whence their 
name, — and of the various media in which waves of sound, heat, 
light, etc., are propagated." — Tail. 



78 DYNAMICS OF A PARTICLE. 

Ex. 1. The motion of the piston of a steam-engine is the 
more nearly harmonic the greater the ratio of the connect- 
ing-vod to the crank axis. (See Fig. 119.) 

2. Combine two S. H. M/s of equal period and which 
take place in the same straight line. 

Ans. Amplitude = (r + r') cos cot, a S. II. M. 

3. Combine two S. H. M/s in directions at right angles to 
each other if the periods are the same. 

[Take them parallel to the axes of X, Y\ then 

x = r cos Got, y = r' cos cot. 

Eliminate / and the locus results. It is x/r — y/r' = 0, a 
straight line.] 

4. From P (Fig. 47) let fall PN perpendicular to OY. 
Show that N has a S. H. M. differing \ in phase from J/ 
but of the same amplitude and period. 

[For 0N= r sin cot = r cos (oot — tt/2).] 

5. Show that a uniform circular motion is equivalent to 
two simultaneous S. H. M/s of equal amplitude and period 
but differing £ in jxhase. 

6. Combine two S. H. M/s in directions at right angles to 
each other if the periods are as 1 to 2. Ans. A parabola. 

These loci and others more complex may be traced out 
mechanically by Blackburn's pendulum, an instrument to 
be found in most physical laboratories. 

73. Simple Pendulum. Consider the motion of a particle 

of mass m suspended from a point by a string of length /, 

Fig. 48 the force of gravity being the acting 

force. The arrangement is known as 

a simple pendulum. 

Let P be the position of the parti- 
cle at any time. Denote the angle 
between OP and the vertical CO by 6. 
The acting force mg may be resolved 
into my sin 6 along the tangent at P 
and my cos 6 at right angles to it. 
The motion is due to the tangential 
component only as the other being equal to the pull of the 
string cannot affect it, 




CONSTRAINED MOTION. 79 

Now accel. along tang. = mg sin 6/m 

= gd, if 6 is small 

= OP x g/l; 

or, the acceleration is proportional to the displacement OP, 
and therefore the motion in the small arc OP is a S. H. M. 
The time t of an oscillation being one-half the period, we 
have 



t = 7i V displac. /accel. 

= 7tVTlg, 

a result independent of the length of the arc. Hence in all 
small arcs the times of oscillation are the same, and the vibra- 
tions of a pendulum are therefore said to be isochronous. 

74. We may now find the length I of a pendulum beat- 
ing seconds at any place. For take a pendulum of length l t 
and count the number ,of oscillations n t in a day. The 
number of seconds in this time is 86,400. But if t, t t are 
the times of oscillation of the two pendulums, we have 

86,400/w, = tjt = VTJI, 

and therefore I is found. 

AVe have now at once the means of computing the value 
of g at the place in question.* Thus 

g=nH. 

Another method of finding g is given in Ex. 6, Art. 55. 
The pendulum method is the more accurate. See also 
Art. 156. 



* In measuring the accelerating force of gravity Galileo was led to 
the invention of the pendulum as a means of measuring small por- 
tions of time. He found g = ol ft. The true value was first found 
by Huygens— who also gave us the pendulum clock. 



80 DYNAMICS OF A PARTICLE. 

75. If a pendulum of length / makes n oscillations in a 
given time r at a place where the acceleration of gravity is 
g, then 

r/n = t = n Vl/g. 

Suppose (1) the length of the jDendulum slightly changed 
by an amount A. By differentiation 

-rdn/n* = %Vl/gldl 

and — dn = n dl/2l — n\/2l, 

giving the loss in the number of oscillations. 

Suppose (2) the pendulum carried to a place where g is 

different by an amount y, the length remaining the same; 

then 

n 

r dn/n 2 = 2 Vl/g\lg 

and dn = n dg/2g = ny/2g, 

giving the gain in the number of oscillations. 

Suppose (3) the pendulum carried to a height h above 
the earth's surface; then, since g varies as 1/r 2 , r being the 
earth's radius, we have 

r/n = cr VI, where c is a constant. 

Hence — r dn/n* = c VI dr, 

— dn = n dr/r = nli/r, 

giving the loss in the number of oscillations. 

Similarly if carried to a depth h below the surface the 
gain in the number of oscillations would be nh/r. 

Ex. 1. If a pendulum, length /, vibrates n times in s sec- 
onds, prove 

Z7rV = gs 2 . 

2. Find the number of oscillations made by a pendulum 
a yard long in one minute. Ans. 62.57. 



CONSTRAINED MOTION. 81 

3. A particle attached to a fine wire vibrates 60 times in 
3 minutes: find the length of the wire. Ans. 29.36 ft. 

4. A seconds pendulum makes 10 oscillations more in 24 
hours at the foot of a mountain than at the top : find the 
height h of the mountain. Ans. | mile nearly. 

5. A pendulum which beats seconds at the surface when 
carried to the bottom of a mine gains 5 beats in 24 hours: 
find the depth of the mine. 

6. At New York the value of g is 32.16 ft. : find the 
length of the seconds pendulum there. Ans. 39.1 in. 

7. A clock gains 30 m. per week : how much should the 
pendulum be shortened for correct time. 

Ans. 0.006 of its length. 

8. A clock keeps correct time at a place where g is 
32.24 ft. : show that it will gain 3 m. 20 sec. per day at a 
place where g is 32.09 ft. 

76. The problem of the pendulum is a special case of 
the motion of a particle constrained to move on a smooth 
vertical circle under the action of 
gravity. 

Take the origin at O, the lowest 
point of the circle; OX horizon- 
tal, O Y vertical, and let x, y denote 
the co-ordinates of P any position 
of the particle. Let A be its initial 
position, AP = s, and OOP = 6, 
OCA = (3, both expressed in cir- 
cular measure. Draw A B, PD parallel to OX. 

Then for an indefinitely small distance ds, the path may 
be regarded as a straight line and the general equation of 
motion is 

d*s/df = - g sin 6. 
Integrating, we find 

v 1 — (ds/dty — %gr (cos 6 — cos (3) = 2g x BD, 
or the velocity at any point P is the same as that acquired 



Fig 


49 




Y 






I C 


^ 




\ D 


X 




y r 



82 



DYNAMICS OF A PARTICLE. 



in falling through the height BD, that is in falling from 
the horizontal line AB* 

To find the time of motion from A to P. We have 
ds = rdO, and therefore, from the preceding equation, 

(dd/dty = 2rj (cos 6 - cos 0)/r, 

/o 
dd/ Vcos 6 - cos ft, 

the — sign being taken because 6 decreases as t increases. 
This equation cannot be integrated by the ordinary methods. 
If, however, ft is so small that powers above the second may 
be neglected, we have cos 6 — cos ft = {ft 2 — 6 2 )/'2, and 

which gives the value of t. 

If = 0, we get the time of reaching the lowest point to 

be — Vi'~/g,the result already found in Art. 71 for the 
pendulum. 
The pressure N of the particle on the circle at P is due 

* The following is Galileo's experimental proof of this principle. 
"In Fig. 50 let a string AG with a weight C appended be fastened 
to a point A in the vertical plane 



Fig. 50 



ACB, so that the weight may 
swing in the circular arc CBD. 
If the weight be let fall from 1) 
it will descend to Baud rise again 
to C, the velocity at the lowest 
point i? acquired by falling down 
BB exactly sufficing to carry it 
up to the horizontal line from 
which it fell. Now let a nail be 
fixed at E in the vertical line 
AB so that on the side of B the 
weight maybe compelled to move 
in the circular arc GB of which 
the centre is E. Then G being 
in the horizontal line CB, let the weight fall from G and it will be 
found that it still rises exactly to C before its velocity is extin- 
guished . "— Whewell. 




CONSTRAINED MOTION. 



83 



both to the centrifugal force and to the weight of the par- 
ticle. Hence if m is the mass of the particle, 

N = mv 2 /r -f- mg cos 0. 

Ex. 1. Compare the times of a particle sliding clown a 
small arc A of a vertical circle to the time of sliding down 
the chord A 0. Ans. n : 4. 

2. A particle starts from the highest point of a smooth 
vertical circle and slides down the convex side under the 
action of gravity: find where it leaves the circle. 

Ans. At a depth = radius/3. 

3. If particles start from the common highest point of a 
series of vertical circles down their convex sides, find the 
locus of the points of departure from the circles. 

Ans. A straight line. 

4. A mass m hanging at the end of a string of length I 
is projected with a velocity u so as to describe a vertical 
circle. Show that the tension of the string T and the 
velocity v at any point in the path whose vertical height is 
h are found from 

u* = v 2 + 2g7i ; 
Tl/m = it 2 + g{l - U). 

Hence show that if u 2 > 5gl the particle will perform com- 
plete revolutions ; if u 2 < 2gl it will oscillate in an arc less 
than a semi-circumference; and if u 2 > %gl and< bgl it will 
cease to describe a circle and the 
motion become parabolic. 

77. Centrifugal Pendulum. — 
Suppose a particle of mass m sus- 
pended by a string from a point 
and caused to swing about the vert- 
ical axis OA with a uniform velo- 
city v in a circular path. Such 
an arrangement is called a centrif- 
ugal pendulum. 

Let B be the position of the par- 
ticle at any time. Denote the angle 
between OB and the vertical OA by 6. 




84 DYNAMICS OF A PARTICLE. 

The particle is acted on by two forces, the gravity force 
my and the pull P in the string directed to the point 0. 
Since the resultant motion is the same as in the case of a 
particle acted on by a centripetal force (directed to A), it 
follows that the resultant of the acting forces must be di- 
rected along the radius BA (= r) and form a centripetal 
force. Hence, there being no resultant vertical force, we 
have 

P cos 6 — mg = 0; 

centrip. force = P sin 6 — mg tan 6; 
and the acceleration a due to the centripetal force is thus 
g tan 6. 

But since v is the velocity in the circular path, 

a = v^/r. 
Eliminating a, 

v* = rg tan 8, 

or liv* = gr 2 , 

if we put the height OA — li ; which gives the relation 

between h, v, r. 

The time T in which the particle makes a revolution, or 
the period, is given by 

T — circum. of path/velocity 
= 2nr/v 
— 2tt Vh/g seconds. 

Also, we may determine h so that the number of revolu- 
tions per second may be any desired number, n for example. 
Then 

v = %nr x n. 

But v Vh — r Vg. 

Whence eliminating v, and putting g = 32.2, 

Jui* = 0.815 feet, 
the relation required. 



CONSTRAINED MOTION. 



85 



Ex. In going round a ring 100 ft. in diameter on a 50- 
in. bicycle a velocity of 10 ft. per sec. is attained. Find 
the distance of the highest point of the wheel from the vert- 
ical through the lowest point. Ans. 3 in. nearly. 

78. The centrifugal pendulum may be used as a regu- 
lator of mechanical motion. An apparatus of this kind, 
known as the Governor, was applied by James Watt to the 
steam-engine.* 

In Fig. 52, which represents a 
Corliss-engine governor, as the 
speed of the engine increases 
the spindle, A revolves more 
quickly, and the balls separate ; 
as it diminishes, the balls come 
together. The slide rises and 
falls accordingly, and by means 
of a set of levers C the steam- 
valves of the engine are acted 
on, and the supply of steam ad- 
mitted to the cylinder regu- 
lated. 

Ex. 1. Explain clearly the 
pendulum motion in the gover- 
nor. 

2. Does the Watt governor 
prevent increase of speed ? 

3. Find the length of a Watt governor that will run GO 
revolutions per minute. Ans. 9.78 in. 




* "If a pair of common fire-tongs suspended by a cord from the 
top be made to turn by the twisting or untwisting- of the cord the 
legs will separate from each other with force proportioned to the 
speed of rotation. Mr. Watt adapted this fact most ingeniously to 
the regulation of the speed of his steam-engine." 



CHAPTEH IV. 
STATICS OF A RIGID BODY. 

79. Having studied the behavior of a particle under the 
action of forces, we proceed to study the behavior of a body 
of finite size, a body being regarded as a collection of par- 
ticles. 

The directions of the forces applied to a particle must 
necessarily all pass through the particle. In a body the 
directions need not all pass through one point. Besides, 
forces applied to a body may cause it to change its shape 
as well as to change its position. To exclude the former, 
we shall for the present assume that the body cannot be 
made to change its form or be distorted by the action of 
the forces applied. To such a body the name of Rigid Body 
is given. Though bodies differ more or less as regards 
rigidity, we are not acquainted with one perfectly rigid; so 
that a result deduced on the hypothesis of a body being 
perfectly rigid can only be regarded as an approximation to 
the actual state of the case in practice. The hypothesis is 
made only for convenience of study, as it is simpler to dis- 
cuss the properties of bodies one at a time than to attempt 
to grasp all at one time. 

As in particle motion, the first step will be to combine 
the acting forces, all of which are supposed to lie in the 
same plane. 

80. Composition of Forces. — Suppose forces F x , F 2 , F a , 

F A in one plane to act on a body at 

Fi 9-"/ F3 different points A, B, C, D. Each 

ScK / particle acts on the particle next it, 

f. i \)<^J- and is acted on by it in return. These 

— *4-£ ~\ ) internal forces forming actions and 

l~ \f reactions occur always in pairs, and 

— \ J being equal in magnitude and opposite 

in sense, are themselves in equilibrium. 



COMPOSITION OF FORCES. 



87 




Hence we need only consider the external 
forces F x , F n , F 2 , F i so far as the motion 
of the body is concerned. 

Prolong the directions of all the forces, 
and suppose these directions to meet in 
a common point 0. All of the forces may 
he conceived to act at this point. The 
motion will consequently be the same as if the whole 
body were concentrated into a single particle at and the 
resultant force R would be found graphically as in Art. 47 
by plotting the polygon abcde, whose closing side ae would 
be this resultant in magnitude and direction; or by analysis, 
as in Art. 49, by resolving the forces along tAvo axes through 
in directions at right angles to each other, and making 
the sums of the components in each direction equal to 



81. If the directions of the forces do not all meet in 
a point, Ave can still find the resultant by Fig. 55 

repeated applications of the parallelogram 
of forces. For the resultant of F 1 at A 
and F 2 at B is the resultant B 1 of these 
forces acting at D\ the resultant of R 1 at 
D and F 3 at C is the resultant R 2 of R x 
and F 3 at E. Hence i? 2 is the resultant 
of F l , F„ , F 2 acting at A, B, C respec- 
tively. 

This construction is often incoiwenient. 
modification or rather combination of it and the preceding 
is more practical: Plot to scale the forces F 1} F 2 , F 3 in 
order, the line ad representing F x , be representing F^ , and 
cd representing F 3 . The line ad closing the polygon will 
on the same scale represent the resultant R of the forces in 
magnitude and direction. To pro\'e this, join ae. Then<76* 
is. the resultant of ab 9 be, and ad is the resultant of ae, cd, 




The following 



STATICS OF A RIGID BODY. 



that is, of ab, be, cd. Hence the resultant is determined in 
magnitude and direction. 

To find its position, that is, some point in its line of 
action. At any point p in the line of action of F 1 apply any 
two equal forces R x , R a in opposite directions. Let R 3 along 
qp be the resultant of F 1 and R x ; R t along rq of F^ and R 3 ; 
and R b of F z and R A . Hence the two forces i? 2 , R b are equiva- 

Fig. 56 





lent to the three forces F iy F iy F 3 . The resultant R of i? 2 , 
R b passes through the point t, in which their directions inter- 
sect. Hence if through t a line equal and parallel to ad be 
drawn, it will represent the resultant R of F x , F i} F 3 in 
magnitude, direction, and position. 

Again, since ah represents the force F x , if we draw aO,bO 
parallel to R x , R 3 , the sides of the triangle Oab will represent 
the forces R x ,F lf R 3 acting at p. Similarly, the sides of the 
triangles Obc, Ocd will represent the forces at q, r. But R x 
is any force. Hence Oa is any line, and the position of the 
point is arbitrary. The point is called the Pole. 

This gives us the key to the following rule for finding 
graphically the resultant of any number of forces acting in 
one plane on a rigid body. Construct a polygon abed to 
scale, whose sides are parallel to and in the same sense as 
the forces ; the closing side will represent the resultant in 
magnitude and direction. 



COMPOSITION OF FORCES. 



89 



From any point 0, draw lines Oa, Ob, Oc, Od to the angu- 
lar points of the force polygon. From any point p in F x 
draw pq parallel to Ob to meet F 2 in q, and draw qr paral- 
lel to Oc. A line through t, the intersection of pt parallel 
to Oa, and rt parallel to Od, will give the resultant in posi- 
tion. Hence it is completely determined. 

82. It is* evident that a force equal and opposite to the 
resultant R would hold the forces F x , F 2 , F 3 in equilib- 
rium. Hence forces in equilibrium in a plane may be rep- 
resented by the sides of a closed polygon abed, whose sides 
are parallel and in the same sense as the forces. 

The converse of this, that if forces acting in a plane can 
be represented by the sides of a closed polygon which are 
parallel to and in the same sense as the forces, they are in 
equilibrium, is not true. For the polygon would be the 
same, no matter what the position of the forces may be. 
This condition, in fact, provides against translation only. 
An additional condition to provide against rotation is neces- 
sary. (See Art. 88.) 

Ex. 1. Three forces P, Q, R are represented in direc- 
tion by the sides of an equilateral triangle taken the same 
way round: show that their resultant is 



V(P 2 + G 2 + R'-FQ-QR- RP). 



Fig. 57 



2. If R is the resultant of two forces P 
and Q, and S the resultant of Pand R, show 
that the resultant of Q and Sis 2R. 

3. In a jib-crane a weight of 20 tons hangs 
at rest: find the pull P of the chain AB if 
A = 2 AB. An*. P = 10 tons. 

4. A square frame has a force 4 acting from 
A to B, 5 from B to C, 6 from A to I), and 
7 from D to C: find the resultant in magni- 
tude and position. 'ww;*. 

5. An interesting application of the triangle 
of forces is afforded by the suspension-bridge 




90 



STATICS OF A RIGID BODY. 



with the roadway uniformly loaded. 
is represented in the figure. 

Fig. 58 



One half of the bridge 





m^X. 






*X 




\ 






'Q 




/ 








L 


/ 








Q— > 


e/ b 




HJ x 







■y, . 

1 

1 
1 




14" wZ 







[Let ^4i? represent the 
pier, AC the suspension 
cable, BD the roadway, and 
AE the anchorage cable. 
The roadway is suspended 
by rods from the cable, and 
the weight on the cable may 
therefore be assumed to be 
uniform per foot length, 
and its direction GH to 
pass through H, the mid- 
dle point of BD. If I is 
the span and w the weight 
per unit of length, the load acting at G is wl/2. The forces 
acting on AC are the weight wl/ 2 and the tensions PQ 
at its ends, which act along the tangents at those points. 
The tangent CO at C is horizontal, and the direction 
GOH of wl/2 is vertical. Hence^ is the direction of P. 
If d is the deflection CL of the span, draw the triangle 
of forces, and show that Q = wF/8cl. 
What is the value of P ? 

Also prove that the curve of the chain A C is a parabola. 
(Take (7 as origin, CL, CO axes; then equation to curve 
will be found to be Idx* = Vy, a parabola.)] 

83. Parallel Forces. — The case of the forces being paral- 
lel is of special importance. Draw ab, be to scale, to rep- 
resent the parallel 
forces F l , F 2 : the 
line ca closing the 
polygon ( in this 
case a straight line) 
will represent a 
force equal and op- 
posite to the re- 
sultant. Hence 
the magnitude of the resultant is equal to the sum of the 
forces. 



X 



COMPOSITION OF FORCES. 



91 



To find its posit ion. Take any point 0, and join Oa, Ob, 
Oc. Draw CA parallel to Oa, AB parallel Ob, B C parallel 
Oc. Then C, the intersection of A C and BC, is a point on 
the resultant. A line through O equal and parallel to ac, 
that is, parallel to F 1 or F 3 , will therefore give the result- 
ant in magnitude, direction, and position. 

We may readily find an expression for the position of the 
point D, in which the resultant cuts AB. From similar 
triangles Oab, A CD; Ocb, BCD] 

CD /AD = ab/Ob, CD/BD = cb/Ob. 

Eliminate CD and Ob, and 

BD/AD = ab/cb = F l /F 2 , 



F 1 X AD = F. 2 X BD; 
which, since the whole distance AB i 
position of D, 

Illustration. — Make an ap- 
paratus as in Fig. 60, and 
find the point A of balanc- 
ing of known weights by 
trial. Compute its position 
and compare results. 

Compare IT with the sum 
of the weights strung along 
the rod. 

1. Draw the figure corre- 
sponding to Fig. 59 when F 1 , 
F 2 act in opposite directions. 



known, giyes the 



r~\ 




to which a hook is attached. 



Show that R= F i - F 2 . 
. 2. A pulley is a wheel or 
sheaf with a groove round its 
outer edge, and capable of re- 
volving freely about an axis 
through its centre 0. This 
axis is fixed in a frame or block 



92 



STATICS OF A RIGID BODY. 



Fig. 62 



Fig. 63 





In a, fixed pulley let a cord passing over the sheaf, sup- 
port a weight IF. The pull F on the string 
.being the same throughout its length, F — W; 
and if P is the pressure on the support, we 
have the pulley acted on by 4 forces, F, W, P, 
and its weight w. Hence for equilibrium 
P = F-{- W+ w = 2 W + w, 
In a movable pulley the block is supported 
by a cord passing under the sheaf. 

3. InFig.63wehave 
a single fixed and 
a single movable 
pulley. If the ropes 
2 and 3 are parallel, and W is the 
weight of the movable pulley, 
prove 

W+ w = 2F. 

Find the pull on the hook at C. 

4. In a pulley tackle the upper 
and lower blocks each contain two 
sheaves, and the same rope passes 
round all: prove 

Fig. 64 

W-\- wt. lower 
block = \F, 

supposing all 

of the cords 

to be parallel. 
5. A weight of 400 lbs is being raised by | 
a pair of double pulley-blocks. The rope 
is fastened to the upper block, and the parts 
of the rope (whose weight may be disregarded) are con- 
sidered vertical. Each block weighs 10 lbs. Find the 
pressure on the axle of the upper block. An*. 522.5 pounds. 

84. Moments. — If, besides acting in opposite directions, 
F i , F 2 are equal, the points a and c in the construction dia- 




MOMENTS. 93 

gram (Fig. 59) coincide, and the resultant is zero. The 
lines AG, BO become parallel, and therefore do not inter- 
sect. 

A consideration of Fig. 65 will show that the F>g- 65 
tendency of the forces is to turn the body round 
an axis. We are thus led to discuss the case of 
forces that cause turning. 

Suppose a force F in the plane of the paper 
acting on a body, and causing it to turn about 
an axis through a point perpendicular to the 

plane of the paper. The turning effect 
^. depends on the magnitude of the force and 

\ on its distance from the axis, and the prod- 

uct of the two may be regarded as a 
~~p *° measure of the importance of the force in 
producing turning. This product is called 
the Moment of the Force about the Point, so 
that we may define the moment of a force 
about a point to be numerically equal to the product of the 
force and the perpendicular let fall from the 'point on its 
line of action. Thus the moment of F about a fixed point 
is Fp, p being the perpendicular let fall from on F. 

The unit of moment depends on the units of force and 
distance, and is named a foot-poundal, a foot-pound, an inch- 
ton, etc., according to the units of force and distance em- 
ployed. (See Art. 130.) 

It is evident that the direction of turning about is as 
indicated by the arrow in the figure. Reverse the direc- 
tion of the force, and the direction of turning is reversed. 
To indicate the sense of the turning it is usual to call the 
moment of a force about a point negative when the ten- 
dency to produce turning is in the direction in which the 
hands of a clock move, and positive when the tendency is 



94 



STATICS OF A RIGID BODY. 



\ F'9- 67 



in the opposite direction. Thus in the figure the moment 
is — Fp. 

85. The moment of a force F with refer- 
ence to a fixed point may be represented 
graph ically. For if ab represent F plotted to 
scale and Oc the distance^, then the moment 
Fp is represented by ab X Oc, that is, by 
twice the area of the triangle Oab, which has 
. ab for base and Oc for altitude. 



Fig. 68 



We may now find the relation between the moment of a 
force AD about a given point O, 
and the moments of its components 
AB, AC about the same point. 
For from geometry 
OAD = OAO + OCD + ADC 
= OAC + OAB; 
that is, the moment of AD about 
to the sum of the moments of AB, AC 




proved more 

Fig. 69 



is equal 
about 0. 

This very important proposition may be 
generally as follows: Let F x , F 2 , . . . be 
the forces acting at A, R their result- 
ant, and any point in the plane of the 
forces. From let fall the perpendic- 
ulars Oa, Ob, ... Ol(mF x ,F 2f ... R. 
Join A 0. Then, since the component 
of R in any direction is equal to the 
sum of the components of F y , F Q , . . . 
in the same direction (Art. 49), take the direction A Fat 
right angles to A 0, and we have 

Rsm OAl = F x sin OAa + F 2 sin OAb + . . ., (1) 
or R x 01 = F, X Oa + F, X Ob + . . .; (2) 




MOMEKTS. 95 

which shows that the moment of the force R about is 
equal to the sum of the moments of its component forces 
about the scone point. 

Two important consequences follow : 

(a) If the direction of R is reversed, the forces F l9 F 9i 
. . . — R are in equilibrium, and we have 

= R X 01 + F x X Oa + F 2 x Ob -f . . .5 

or, ?#7/ew forces acting at a point are in equilibrium, the 
algebraic sum of their moments about any point in their 
plane is zero. 

(b) When forces acting at a point A are in equilibrium, 
their moment about any point gives the same relation 
as the resolution of the forces about an axis through A 
perpendicular to A 0. One equation is a multiple of the 
other, equation (2) being deduced from (1) by multiplying 
by A 0. See Art. 100 for an illustration of this. 

Ex. 1. Find the moment of a force about any point in 
its line of action. 

2. Compare in magnitude and direction the moments of 
two forces about any point on their resultant. 

2a. Henee find the algebraic sum of the moments of 
two or more forces about any point situated on their re- 
sultant. 

do Is there any reason why a man should put his shoulder 
to the spoke rather than to the body of a wagon in helping 
it up hill ? 

4. If R is the resultant of two parallel 
forces F xi F 2 , and able is any line per- 
pendicular to their lines of action, prove 
independently of Art. 85 that 

R x al = F x Xab + F 2 X ac. 
[Follows from F x X bl ' = F^ X cl ; F 
Fy + &* = & Art. 83.] 

5. Show that the dimensions of the 
moment of a force about a point are MLr/T*. 



Fig. 70 



96 



STATICS OF A KIGID BODY. 



86. Couples.- 

Fig. 71 



-Let ns return to the two equal and op- 
posite parallel forces, Art. 84. The 
moments of the forces about any point 
a are — F X ab and + F X ac, respect- 

ively. Hence the measure of the 

turning effect of the two forces would 
be — F X ab + F X ac or F X be, 
that is, the product of one of the forces 
by the distance between the directions of the forces. 

To the system of two equal parallel forces acting in op- 
posite directions but not in the same straight line the 
name Couple is given. The line be being the distance be- 
tween the lines of action of the 
forces is called the Arm of the 
couple, and the product Fx be, 
or force X arm, is called the 
Moment of the couple, or the 
Torque. 

An example of a couple is 
seen in the copying-press. The 
handle is pushed at A and 
pulled at B, the push and pull 
forming a couple. In conse- 
quence the screw rotates. 

It requires a couple to wind a watch. 



Fig.72 




87. The moment of a couple depending only on the 
magnitude of the forces and the distance between them, the 
effect of a couple is not altered by turning the arm through 
auy angle about one end, nor by moving the arm parallel to 
itself in the plane of the couple, nor by changing the couple 
into another couple having the same moment. 

It hence follows that the resultant of a number of couples 
in a plane is a couple whose moment is equal to the sum of 
their moments. 



97 



Fig. 73 




It also follows that a single force F and a con pie P, P 
acting on a body cannot be in equi- 
librium. For let the moment of 
the couple be Pa, a being its arm. 
Keplace the couple P, P by a 
couple F, F of arm b, so that Fb 
= Pa, and place it in the plane so 
that one of its forces F is opposite p ,r 
to the single force F. The two 
forces F, F at C are in equilibrium, leaving the single force 
F at D unbalanced. Hence there cannot be equilibrium. 

Ex. 1. Three forces are represented in magnitude, direc- 
tion, and position by the sides of a triangle taken the same 
way round: show that they form a couple whose moment 
is numerically equal to twice the area of the triangle. 

2. Four forces are represented in magnitude, direction, 
and position by the sides of a square taken the same way 
round : prove that they form a couple whose moment is 
numerically equal to twice the area of the square. 

88. Eemembering that the turning effect of a force is 
measured by the moment of the force, we can now find the 
conditions of equilibrium when any number of forces F l9 F 2 , 
... act on a rigid body at different points A, B, . . ., all of 
the forces being in the same plane. For the reasons stated in 
Art. 80, it is necessary to consider the external forces only. 
At any point introduce two forces F l , F', each equal 
to F 1 , and of opposite directions. This will 
not disturb the equilibrium. Now F x and 
F x form a couple of moment F x p x if p x is 
t the distance of from the line of action of 
F x . Thus F x at A is equivalent to F x at 0, 
and a couple F x p x . Treating the other 
forces in the same way, we have the forces 
F x at A, F 2 at B, . . . equivalent to F x , F 9 , 
... at 0, and the couples Fj\ , F„j). 2 ..... 
The forces at may be combined into a single resultant 



Fig. 74 




98 STATICS OF A RIGID BODY. 

R and the couples into a single couple G whose moment is 
equal to the sum of their moments. But a single force 
and a single couple cannot be in equilibrium. Hence for 
equilibrium we must have 

R = 0, G = 0, 

the conditions sought. 

These conditions may be put in a form more convenient 
for computation. Through draw two lines OX, OY, 
forming axes of coordinates, and let each force be resolved 
into components parallel to these axes. Denote the com- 
ponents by X 1 , Y x ; X % , Y 2 ; : ■„ . re- 
Fig- 75 spectively. 

Now if 2X, 2 Y denote the sums of 

theforcesX,, X %9 . . .; Y x , T %s 

along the axes of X and Y, then 

(sxy + (27f = ps 

and R = can only be satisfied by 

2J=0, 2Y=0. 

Also, since the moment of the resultant about any point 
is equal to the sum of the moments of its components 
about the same point, the moment Fj\ must be equal to 
Y x x x — X x y x where x x , y x are the coordinates of A. Hence 
Q = 2 ( Yx — Xy), and if G = 0, we must have 

Z(Yx- Xy) = Q. 

Hence the conditions of equilibrium may be stated — 

(a) Hie sums of the components of the forces along lines 
parallel to each of tiuo rectangular axes drawn through any 
point in their plane is zero. 

(b) The sum of the moments of the forces about any point 
in the plane is zero. 

From these conditions three unknowns may be determ- 
ined, and no more. Hence, in order that a problem of 



COUPLES. 99 

this kind be determinate, the number of unknown forces 
that enter cannot exceed three. An important applica- 
tion of this principle occurs in finding stresses in roof and 
bridge trusses. 

Ex. 1. What are the conditions of equilibrium of two 
forces ? Also of three forces, two of which are parallel ? 

2. Show that parallel forces are in equilibrium when 
the sum of the forces = 0, and the sum of their moments 
about every point in their plane = 0. 

3. In the single movable pulley (Fig. 63) if the ropes are 
inclined at an angle 0, prove W = 2F cos 6/2. 

4. If the sum of the moments of a number of forces 
acting at a point in a plane about each of three points not 
in the same straight line is zero, the forces are in equilib- 
rium. Prove this. 

5. Let AB represent a rigid rod (as a crowbar) turning 
on a fixed support C. Let a force 

i^be applied at A, and let IF be the Fig ' 76 , 

resistance to be balanced at C. Given } R 

the lengths of A C, CB, it is required a cl 



to find the relation between F, W Za 

when in equilibrium. 

[Neglect for the present the weight 
of the rod. Let i^and TFbe vertical, | F w* 

and let R denote the vertical pressure 

at C; then R must be the resultant of J 7 and W, and there- 
fore 

R = F+W. 
Take moments about C, and 

Fx AC'-Wx BC=0. 
Hence the ratios of F, W, R are found. 

The rod AB is known as a lever, the support C the ful- 
crum, and the distances A C, CB the arms of the lever. 
The equation F x AC — W X BC = is sometimes called 
the principle of the lever.*'] 

6. Show by sketch the positions of force, resistance, and 

* The properties of the lever were first given by Archimedes (b.c. 
287-212). 



100 



STATICS OF A RIGID BODY. 



Fig. 77 



fulcrum in the following levers: wheelbarrow, spade, claw- 
hammer, rowboat, pair of scissors, pair of nut-crackers, the 
forearm. 

7. A lever is 2 ft. loug: where must the fulcrum be 
placed that 10 pounds at one end may balance 30 pounds 
at the other end ? 

8. From a pole resting on the shoulders of two men a 
weight W is suspended. It is n times as far from one man 
as from the other; what does each support ? 

Ans. W/(?i + 1), n W/(n + 1). 

9. Find the relation between F 
and W in a bell-crank lever. A and 
B the bell wires, (J the pivot about 
which the lever turns. 

[The directions of F, W, and the 
reaction R of the pivot meet in a 
point 0. Hence take moments 
about C.~] 

10. A pair of nut-crackers is a 
inches in length and a pressure of p 
pounds will crack a nut placed b 
inches from the hinge: what weight 
placed on the nut would crack it ? 

A ns. pa/b pounds. 

11. Show that a single fixed 
pulley is equivalent to a lever with equal arms (Fig. G2). 

12. Two cylinders fastened together move freely on a 
common axis O which is horizontal, A force F acts by a 
cord coiled round the larger cylinder (or 
wheel), and balances a weight W hang- 
ing from a cord coiled round the smaller 
cylinder (or axle). 

[The apparatus is equivalent to a lever 
with unequal arms, the axis correspond- 
ing to the fulcrum of the lever, and the 
radii to the arms. It is called the ivheel 
and axle. 

We have 

F -f- W = pressure on axis, 
F X radius wheel = W X radius axle.] 




Fig. 78 




101 



13. The axle of a capstan is 2 ft. in diameter. If four 
sailors push with a force of 40 pounds each at the ends of 



FlQ.7^ 




spikes 4 ft. long, find the weight of the anchor that is 
lifted. 

14. If the force is transmitted by toothed wheels, the 
teeth work in each other, so that the motion takes place as 

Tig. 80 




if between two circles (called pitch circles) in rolling con- 
tact. We therefore replace the wheels by the pitch 



102 



STATICS OF A KIGID BODY. 



circles, and consider the force to be tangential to these 

Fin ill PlwlflC! of +V./-W ™-.^^~ : • j. a 



circles at the successive points of 
contact. 

[In a winch we have a lever AB 
combined with toothed wheels, and a 
drum round which the rope attached to 
the weight W is moved. Replace the 
toothed wheels by pitch circles, and 
the mechanism in outline with the 
forces acting is as in Fig. 81. We 
have 

FxAC=Kx CH y 




RX HD= WxDE; 
D % x £H __ rad. drum no. teeth in pinion 



' W~ AC X B 



' no. teeth in wheel 



•] 



Fig. 82 



" length arm 

15. Given that the cranks have 18 in. leverage, the gears 

are 4 to 1, the drum 6 in. in diameter, and the capacitv 

with two-man power is 3 tons: find the force exerted by 

each man. J 

89. Centre of Parallel Forces.— We have seen (Art. 83) 
that two parallel forces F l9 F 2 
acting at A, B are equivalent to 
a single force F % + F^ acting at a 
point D on the line AB, such that 
F 1 x AD = F 2 x BD. 

Similarly, if F 3 is a third force 
acting at C, the three forces are 
equivalent to F x -f F 2 at D and F 3 
at C, or to a single force 
F i+^ t + F % at G, such that 

(F r + F a ) x DG = F 3 x CG, 
and so on. 

This result is entirely independent of the directions of 
the forces, so that the point G will be in the same position 
if the forces are turned in the same sense through the same 



f/ 






A 


D 


\ B 



G\ 



N c 



CEKTRE OF PARALLEL EOECES. 103 

angle about A, B, so as to remain parallel to one another. 
The fixed point G being the centre of the points of applica- 
tion of the parallel forces, is called the Centre of Parallel 
Forces. 

It is convenient to express the coordinates x, y of G, 
the centre of parallel forces, in terms of the coordinates 
x l9 y x ; x i9 y 2 ; . . . of A, B, . . . referred to axes OX, 
OY drawn through, any point in the plane of the forces. 

Since G is in the same position no matter what the 
directions of the forces may be, let them be parallel to OY. 
Take moments about 0, and 

(F, + F 3 + F z )x = F x x x + F& + >A* 

and x is found. 

Next take them parallel to OX, and take moments about 
0; then 

(Fi + F% + F$y-F iyi + Fj % + F s y s , 

and y is found. 

The values of x, y may be written for any number of 
forces, 

x = 2Fx/2F, y = 2Fy/2F, 

when 2 is the symbol of summation. 

It is evident that the same reasoning would apply if the 
points A, By . . . were not in the same plane, the forces 
still being parallel. If z x , z 2 denote their distances from 
a fixed plane, the distance z of G from this plane would be 
given by 

1 = 2 Fz/2F. 

90. Centre of Gravity. — As an illustration of parallel 
forces, consider the force of gravity acting on the particles 



104 



STATICS OF A RIGID BODY. 



Fig. 83 




m,0t <r v™ 2 ! 



of a body. A body may be regarded as built up of parti- 
cles, the weights of the particles forming a system of 
forces whose lines of action, pass- 
ing through the earth's centre, 
are so nearly parallel in a body of 
ordinary size that we may con- 
sider them to be so. If m 1 , m % , 
. . . lbs. are the masses of the 
particles, the parallel forces act- 
ing downwards on them are 
m/j, m a g , . . . poundals, and 
the resultant force would be 
found by adding the forces. To 
0, the centre of these parallel forces, the name of Centre 
of Gravity* is given. Its distance z from the plane of 
X, Y may be written z = 2mgz/2mg. This may also be 
written 

z = 2mz/2m, 
and hence the centre of gravity is also called the centre of 
mass, or by some the centra id. 

91. That the line of action of the resultant force of 
gravity passes through the centre of gravity in all positions 
of a body, suggests an experimental method of finding the 
centre of gravity. Thus conceive the 
body suspended by a string from a point 
P. The forces acting are the resultant 
force of gravity at the centre of gravity, and 
the tension (pull) of the string. The lines 
of action of these forces must lie in the 
vertical through P. Hence, to find G, 
suspend from P, and strike the vertical 
PH; next suspend from any other point 
Q, and strike the vertical QK\ the point 

* The idea of the centre of gravity of bodies is due to Archimedes. 




CENTRE OE PARALLEL FORCES. 105 

of intersection of PR and QK will be the centre of gravity 
G required. 

92. When the particles of a body are so distributed that 
there are always the same number in the same volume, the 
body is said to be of uniform density. The unit of density 
is taken to be the mass of unit volume. Thus, if m is the 
mass and V the volume, then the density p = m/ V. 

If the density is not uniform, we must estimate it at 
every point. Thus for an indefinitely small mass dm of vol- 
ume dv about a point, the density at the point is dm/dv. 

We shall consider bodies of uniform density only. 

Ex. 1. The 0. of G. of a uniform straight rod is at its 
middle point. 

[The rod being uniform, is such that the number of 
particles on one side of the centre C is 
equal to that on the other side, and the 
0. of G-. of every pair being at C, the 0. 
of G-. of the whole is at C] 

2. Find the C. of G. of a triangular 
lamina of uniform thickness and density. 

[Conceive it divided into strips by lines 
parallel to AB. The C. of G. of each 
strip will be at its middle, point. Hence 
the 0. of G. of the whole will lie on the 
line CD joining C to D, the middle point of A B. Simi- 
larly, it will lie on the line joining A to E, the middle point 
of BO. Hence it is at G, the intersection of CD and AE. 

Join BE. The triangles DGE, AGC are similar, and 
DE = \A C. . • . DG = \CG = ±CD.] 

3. Prove that the C. of G. of a triangle (triangular 
lamina, strictly) coincides with that of three equal weights 
placed at its angular points. 

4. Three men support a heavy triangular board at its 
corners: compare the weights supported by each man. 

5. The sides of a triangle are 3, 4, 5 : find the distance of 
the C. of G. from the angles. 

Ans. Vld/B, 2 i/U/S, 5/3. 

6. Explain how an instrument which stands on three 




106 



STATICS OF A RIGID BODY. 



screws can be " levelled " by means of these screws. Take 
a galvanometer, for example. 



Fig. 



mfi Y 



93. We pass at once to the 
centre of gravity of a system of 
bodies rigidly connected, by con- 
sidering that each body may be 
conceived concentrated into a par. 
tide of equal mass acting at the 
centre of gravity of the body. 
Thus, reasoning as in Art. 89, if 
m l9 m a , . . . represent the masses 
of two bodies, the centre of gravity 



G of the system will be found from 

m 1 X G X G = m 2 x G 2 G. 

Similarly, if G, G 1 are given, G 2 may be 
found. 

Ex. 1. Weights of 1, 2, 3, 4, 5 pounds are 
strung on a uniform rod AB, whose weight is 
3 pounds, at distances of 4 in. from each other: 
find the point at which the rod will balance. 
Ans. AG = 10§ in. 

2. Find the C. of G. of a T-iron, depth 
= d; depth of web = d x \ breadth of flange = b; 
breadth of web = b 1 



Fig. 87 

ic 



Ans. CG{btl-dSb 



Fig. 88 



«(«-$)• 




3. A common form of cross-sec- 
tion of a reservoir wall or embank- 
ment wall is a trapezoid whose top 
and bottom sides are parallel. If 
top side = a, bottom = b, and 
height = h, show that 

-_h P a + b\ 



■ \a + b I' 



CENTRE OF PARALLEL FORCES. 



107 



Fig. 
c 




4. Hence show, and also show independently, that if 
one -fourth part of a 
triangle is cut off by a F 
line parallel to the base, —-"—-- 
the 0. of G-. of the re- 
mainder is at 2/9 of the 
line joining the vertex 
to the middle point of the base. 

5. From a circular disc another circular disc described 
on its radius as diameter is cut : show 

d that the C. of G. is distant 1/6 radius 
from the centre. 

6. Prove the following construction 

for finding the 0. of G-. of a trapezoid 

ABCD. (Fig. 89.) Prolong AB, making 

BE — CD; prolong DC, making CF = 

AB. The point of intersection of EF 

and HK, the line joining the middle 

points of AB, CD, is the C. of G. 

7. Prove the following rule for finding the 0. of G. of a 

quadrilateral: (Fig. 90.) Draw the diagonals. MakeJi^ = 

DE. The 0. of G. of the triangle CFB is also that of the 

quadrilateral. 

94. The application of the general formula of Art. 90 to 
finding the C. of G. of bodies may be further illustrated by 
the following examples: 

Ex. 1. To find the C. of G. of a circular arc AB. 

[The C. of G. must lie on a line OCX 
joining the centre to the middle point 
C of the arc. Let ds denote the length 
of the indefinitely small part PQ of the 
arc, h its cross-section, p its density. 
The mass of PQ = kpds. The coordi- 
nates of the C. of G.of PQ are the same 
as those of P, or x, y. Hence 

x{=0G) = fkpds X x/fkpds 
= fxds/fds. 
If COP = 6 and OP = r, then 

x = r cos 6, ds = rdO, and x =fr cos Odd j\W. 



Fig. 91 




108 



STATICS OF A RIGID BODY. 



If angle A OB = 2/?, then, integrating between the limits 

+ A - A 

x = r sin/?//?.] 

2. To find the C. of G. of a semicircle. Ans. x = 2r/7t. 

3. To find C. of G. of a circular sector ^4 OB if angle 
A OB ==2/3. (See last figure. ) 

[Divide the sector into triangular pieces as OPQ. Area 
OPQ = \rdd xr= \rd6. If G, is C. of G. of OP Q, then 

x— OM= 0G X cos 6 = fr cos 0. 

/' + £ /* + 

fr cos X irV/0/ / £rW = fr sin/?//?.] 

4. Find the C. of G. of a circular ring, radius OA = r, , 
06' = r 2 , angle v4 0i? = 2 /?. 

[A practical application would be 
the front surface of a circular arch 
of which r, is the radius of intrados 
and r a radius of extrados. ] 
Ans.lc==%(r t a —r 1 *)sm/3/Jr*--r 1 *)/3. 

5. For a semicircle, a; = 4r/37r; 
for a quadrant of a circle, x = 
4 V2r/37t. 

95. Stability. — If a body suspended from 
slightly displaced from its position of equi- 
librium and let go, it will turn about the 
point of support. If the centre of gravity 
G is belotc the point of support, the ten- 
dency is for the body to return to its original 
position. This may perhaps be made more 
evident by resolving the weight W into F x 
along GO and F 2 at right angles to F x . 
The force F l gives the pressure on the sup- 
port, and i^ 2 the turning force which tends to 
swing the body to its former position. In 
this case the body in its original position is 
said to be in stable equilibrium. 




point be 

Fig. 93 





STABILITY. 1 09 

In Fig. 94 the centre of gravity is above the point of sup- 
port, and the tendency for the body when Fig. 94 
disturbed is to move farther from its origin- 
al position. In this case the equilibrium 
is unstable. If, however, the point of sup- 
port is at the centre of gravity G, the body 
will remain at rest in any position, and the 
equilibrium is neutral. 

Similarly, if the body, instead of being suspended from a 
point, rests at a point on a fixed support the equilibrium 
may be stable, unstable, or neutral, according as the forces 
acting on the body in its displaced position tend to restore 
it to its original position or make it move farther from 
that position, or are in equilibrium. 

When one point of a body is fixed the resultant of the 
external forces tending to cause translation is balanced by 
the reaction of the constraint at 0. Also, if there is equi- 
librium the tendency to 'turn about in one direction 
must be balanced by the tendency to turn in the opposite 
direction, or the sum of the moments about must be 
zero. This, therefore, is the condition of equilibrium of a 
system of forces acting on a body with one point fixed. 

Ex. 1. In suburban passenger traffic the train must stop 
and start quickly. The engine is built with a large wheel 
base. Why ? 

2. A circular table weighing w lbs. has three equal legs 
at equidistant points on its circumference. The table is 
placed on a level floor. Neglecting the weight of the legs, 
find the smallest weight which, when placed on the table, 
will upset it. Ans. w lbs. 

3. If the table has four legs at equidistant points, find 
the least weight that will upset it. 

4. Suppose A BCD (Fig. 95) to be the cross-section of a 
wall built to withstand the pressure of earth or water on one 
side: for example, the wall of a reservoir or of a railroad 
embankment. Such a wall is called a retaining-wall. 



110 



STATICS OF A RIGID BODY. 





Fig. 95 






C 


D 


E^- 


•"m 


•G 




^A 


H 




B 






W 





[We assume that the wall is built so that it cannot give 

way except in one mass, 
and by being overturned 
about the edge A. Let 
P be the resultant of the 
' forces pressing on the side 
BD, and let be its point 
of application. The weight 
W acts vertically down- 
ward through the centre 
of gravity G. The stability 
depends on the difference 
of Px ^^and W X AH, 
the first tending to over- 
turn, the second to restore. Hence, when the wall is just on 
the point of turning, we have the relation 

W X AH= P X AK] 

5. The centre of gravity of a ladder weighing 50 lbs. is 
12 ft. from one end, which is fixed. What force must a 
man apply at a distance of 6 ft. from this end to raise the 
ladder to a vertical position? A us. 100 pounds. 

6. Find the proper elevation BE of the outer rail on 
a railroad track for a given velocity v of engine weighing 
m lbs., and on a curve of radius r, 

in order that there may be no 
flange or lateral pressure on the 
rails. 

[The forces acting are the weight 
mg of the engine, the centrifugal 
force C(= mv a /r) and the reac- 
tions N l9 jV" 2 of the rails. Since 
there is no flange pressure the reac- 
tions are perpendicular to the track 
AB. Let 6 be the inclination of 
AB to the horizon. 

Eesolving the forces along AB, 
we have 

C cos 6 = mg sin 6, or tan 6 = C/mg = v*/gr. 




BEA3I SUPPORTED AT ONE POIXT. 



Ill 



But if 6 is small, tan 6 = BE J ' AB = elevation/gauge; 

. • . elevation of rail = gauge X v'/gr. 
For standard gauge of 4 ft. 8J in., this gives 

elevation of rail = 7v a /4r inches, nearly, 

v being expressed in feet per second, and r in feet.] 

7. If in (6) the velocity V is expressed in miles per hour, 
show that 

elevation of rail = 15 F 2 /4r inches, nearly, 

the radius r being, as before, expressed in feet. 

8. Find the greatest velocity v a locomotive can have to 
be just on the point of overturning on a curved level track 
of radius r ft., the centre of gravity of the locomotive being 
6 ft. above the rails, and the gauge of the track 4 ft. 8|- in. 

Ans. 3.55 Vr ft. per sec. nearly. 

96. Beam Supported at One Point. — Suppose a beam of 
wood or metal of length %a suspended at a point 0. The 
pressure on the support berng equal to the weight W of the 
beam, the centre of gravity G of the beam will lie in the 
vertical OG. A weight P placed on the beam or suspended 



Fig. 98 



Fig 




from it anywhere except in the line OG will cause the 
beam to take an inclined position, as in Fig. 9S. Suppose 
P suspended at A, where AO — a. The forces P and W 
are parallel, and therefore their resultant must be equal to 
their sum. Being balanced by the reaction of the support 



112 



STATICS OF A RIGID BODY. 



Fig. 99 



0, this resultant must pass through 0. Hence the moment 
of W about is equal to that of P about 0, or 

P X 00 =W X GH, 
or P X a cos 6 = W X h sin 0, 

or P = tan 6. 

a 

By attaching a pointer to the beam free to move over a 
graduated arc we have a means of compar- 
ing weights. An example is afforded by 
the common letter-scale (Fig. 99). 

97. Balance. — If in the beam repre- 
sented in Fig. 97 we place two equal 
weights P, P at the same distance A from 
0, equilibrium will not be disturbed. For 
the moment about is the same, being 
= Pa in both cases. Hence the beam may 
be used for comparing equal weights. At- 
taching pans to A, B, the weights for com- 
parison may be placed in these pans and 
the operation facilitated. Such an arrange- 
ment is the common Balance. 

The parts of a balance should be arranged so as to secure 

the greatest accuracy 

in making the com- F «g- ioi 

parisons, and with 

the least loss of time. 

How can this be 

done ? 

Suppose the 

weights P, Q to be 

unequal, and that 

the points of support 

A, B are equidistant 

from and in the 





BEAM SUPPORTED AT ONE POINT. 113 

same straight line with it. Let AB — la and OG = h. 
Take moments about 0, and 

P X a cos 6 — Wh sin (9 — Qa cos = 0, 

or tan 6= {P - Q)a/W7i. 

Now the balance will indicate small differences P — Q 
the more clearly the greater the angle 6 through which it 
swings for these differences. But tan 6 or 6 is greatest 
when a/ Wh is greatest, that is, when a is large or the beam 
has long arms, when W is small or the beam light, when h 
is small or the centre of gravity is just below the point of 
suspension. Such a balance has great sensibility, and is 
suitable for delicate investigations in Chemistry, Physics, 
Assaying, etc. 

In scales for weighing large masses stability rather than 
sensibility is wanted; that is, for small differences of P and 
Q the angle of deviation of the beam from the horizontal 
as shown by tan 6 should be small. This requires Wh to 
be large or the beam to be heavy, with a long distance be- 
tween the centre of gravity G and the point of suspension 
0. By making the arms long, a balance may be constructed 
which shall possess in a measure both sensibility and stabil- 
ity. As the two conditions are at variance, the amount of 
compromise must be decided by the use to which the bal- 
ance is to be put. 

Ex. 1. If the centre of gravity coincides with the point 
of suspension, the balance is in equilibrium in all positions. 

2. The arms of a balance are equal, but the scale pans 
are not of the same weight. If a body weighs P lbs. in one 
and Q lbs. in the other, find the true weight. 

Ans. i(P+ Q) lbs. 

3. A body placed in one scale pan appears to weigh P 
lbs. and in the other Q lbs. If the pans are of equal 
weight but the arms are not of equal length, show that its 
true weight = VP~Q lbs. 




114 STATICS OF A RIGID BODY. 

98. Steelyard. — Suppose (Fig. 102) a beam suspended 

from a point di- 
ig " l rectly above its centre 

of gravity G. The 
upper edge ABC be- 
ing straight and at 
right angles to 00 
will be horizontal. If 
from the beam we 
suspend two bodies 
of unequal weights P, Q, it will still remain horizontal if 
the moments of P and Q about are equal, or 

P X AO= Qx BO. 
Let the weight P suspended from A be a scale pan. If to 
P we add an unknown weight IT, we shall still have equi- 
librium, provided Q is suspended from a point such that 

(P-f W)AO= Q X CO. 
Subtract these equations, and 

WXA0=QX BC, 
which gives the unknown IF as soon as BC is measured. 

To save measurements of BC at every weighing of a body, 
it is convenient to graduate the beam in the first place. 
Thus suppose P = 1 lb., Q = 2 lbs., and A == 4 in. Then 
OB = 2 in., and a notch can be made at B, which, as the 
weight Q then balances the pan P only, would be marked 0. 

Let now W = 1 lb.; then BC .== -^ X A = 2 in., and C 

would be the position of the 1 lb. mark. Make JF= 2 lbs. 
and BD = 4 in., giving D the 2 lb. mark, and so on. 
Hence in weighing a body it is only necessary to place it in 
the pan and move the weight Q until the notch is found 
where the beam will remain horizontal. The number at 
the notch indicates the weight. This instrument is called 
a Steelyard. 

Ex. 1. Graduate a steelyard to weigh half-pounds. 



BEAM SUPPORTED AT TWO POINTS. 



115 



2. If the point of suspension be not over the centre of 
gravity and the movable weight Q placed at a point H 



Fig. 103 




Fig. 104 



holds the steelyard in a horizontal position, show that 
RB — W X AO/Q, and hence show how to graduate the 
steelyard. 

3. A steelyard beam weighs 3 lbs., the wt. Q is 4 lbs., and 
the distance of the centre of gravity from is 3 in., and of 
the point of suspension of the scale A from (9 5 in. : show 
that the 1 lb. graduation marks are at intervals of f in. 

4. A steelyard weighs IF lbs. and is correctly graduated 
for a movable weight Q: prove that a weight 2Q may be 
used provided a fixed weight IF is suspended at the centre 
of gravity of the steelyard. 

5. A piece is broken off 
the longer arm of a steel- 
yard: show that the customer 
is defrauded. 

99. Beam Supported at 
Two Points. — We naturally 
pass from a body supported 
at a single point to one in 
which two points A and Bare 
supported, as for example 
a beam supported by two 
smooth horizontal pins A 
and /?. The forces acting 
are the weight )Y vertically downwards through and tin* 




116 



STATICS OF A RIGID BODY. 



UN, 



reactions JV", , JV, of the supports .4, i? in directions which 
are unknown. If we assume that they are in the same 
plane, the forces being three in 
Fig. 105 number must meet in some point 

in the vertical through G. 
Until we are able to fix the 
directions of N x or N i9 the 
pz» problem is indeterminate, as 
there is an indefinitely great 
number of points in this line. 
100. If, for example, the beam 
rests on two props A, B in the same horizontal plane, the 
reactions of the props are 
vertical. Hence, resolving 
vertically, and taking mo- 
ments about G, 

from which iV, , JV 2 are 
found. 

Or take moments about A 
and B in succession, and 

N t X AB-W X BC= 0, 
N, x AB-W X AC = 0, 
from which the same values 
result as before. 

The pressures N y , JV a on 
the supports may also be 
determined graphically by 
the method of Art. 81. 

Suppose the beam AB to 
carry besides its own weight 

W a load W % at C, Draw 
a b, be to scale to represent 

W l9 W, and parallel to their directions. The force R 









Fig. 


106 








N, 


c 






N 2 






C 


G 




p 


A 


8 


W, , 


W 
r 


B 


8 




r \ 




a 










I 

b 














c 















SCALE OF DIMENSIONS 



t SCALE 



BEAM SUPPOETED AT TWO POINTS. 117 

represented by the line cba which closes the polygon will 
hold W 1 , W in equilibrium. 

Take any pole and join Oa, Ob, Oc. From any point p 
in the direction of jVj draw pq parallel to aO; draw qr 
parallel to bO and rs parallel to cO. The intersection t 
of pq and sr gives the position of the resultant R. Its 
direction is parallel to cba, and is therefore vertical. 

Join sp and draw 01 parallel to sp. Now cba (or 7?) may 
be resolved into Oa along pq and cO along sr. Also 0« may 
be resolved into 01 along sp and la along ^4. And cO may 
he resolved into cl along si? and /0 along ps. But the forces 
0/, 10 being equal and in opposite directions, must be in 
equilibrium. Hence cba. (or R) is equivalent to la at A and cl 
at 1?. But cba holds IF^ , IF in equilibrium. Hence la, cl hold 
W lf IF in equilibrium, and are the reactions N l3 iV 2 at ^1 and i>. 

Hence the rule, 

(a) Form the force polygon by laying off the forces to scale. 

(b) Select a convenient pole and form the polygon pqrs. 

(c) Draw 01 parallel to the closing line sp dividing ca into 
parts la, cl, which will represent the reactions N y , iV 2 at A 
and B. 

Ex. 1. A beam of 20 ft. span carries a weight of 10 tons 
8 ft. from one end : find the pressures on the end supports. 

2. A highway bridge 25 ft. long weighs 6 tons: find the 
pressures on the abutments when a 2^-ton wagon is \ of the 
distance across. Ans. 5 tons; 3.5 tons. 

3. A beam of 40 ft. span weighs 1 ton per running foot. 
One half of it carries a uniform load (as a train of coal cars) 
of 2 tons and the other of 3 tons per running foot: find 
the pressures on the end supports. Ans. G5 tons; 75 tons. 

4. A truss of 60 

ft. span and weigh- Fig- 107 

ing 100 tons car- § § § 8 g 

ries an Erie con- « « Si « 3 

solidation engine a — in , ^ ■ .ft ..ft. r ft — -^ft * 



as in the figure: 

find the pressures on the supports. 

Ans. GG.C tons; 84.9 tons. 



118 



STATICS OF A RIGID BODY. 



101. Two Beams Hinged. — Again, suppose to the beam 

AB an equal beam 
BC attached at B 
by a hinge or pin, 
and the two beams 
to be supported by 
two pins at A, C 
in the same hori- 
zontal plane. The 
pin B being in 
equilibrium, the re- 
action K of AB on 
it is equal and op- 
posite to that of 
BC on it. Since 
the beams are equal 
the direction of N 
must be horizontal. The weight TFof the beam AB acts 
vertically through the middle point G. Hence the reaction 
N 1 of the pin A must pass through H, the intersection of 
the directions of iVand W. Similarly for the beam BC. 

The forces acting on the beam AB are parallel to the 
sides of the triangle A FIK, and N, N^ may be scaled off 
directly if UK be taken to represent the weight W. 

Put the length AB—l, the height BO = h, the span 
A C = 2a, and we have 

N = W X AK/EH = Wa/27i, 




W=WX AH/KH= WVa* + 4J?/Wi, 



which give the algebraic values of the reactions. 

These values may also be found by taking moments about 



TWO BEAMS HINGED. 119 

A and B in succession. Thus taking moments about A. 

N X HE = W X A K, 

giving iVas before. Similarly for the value of N t . 

The reaction X x may also be found by resolving it into 
two components, X horizontal and Y vertical. Then 

Y=W, X = N=Wa/%h, 



and AT = \/X -j- Y 2 = W iV + 4th */2h, 

as before. 

102. The reaction A^ at A may be resolved into two 
components — a longitudinal force AL along the beam AB 
and a transverse force AE perpendicular to it; also iV^at B 
into a longitudinal force BL and a transverse force BF. 
The longitudinal force diminishes from AL at A to BL at 
B. The transverse forces- AE, BF being each equal to 
HL, are equal to one another. Also HL — HG cos 6 = 

W 

■g- cos 6, so that the transverse forces are equal to half the 

component of IF at right angles to the beam. 

The values of the transverse forces are the greater the 
smaller the angle 6 is, or the more nearly the beam is hori- 
zontal. When the beam is horizontal and the external 
forces are vertical, the longitudinal forces disappear and 
the transverse forces alone enter. In the study of struc- 
tures it is necessary to consider the effect of both of these 
classes of forces. 

103. The computation of the longitudinal and trans- 
verse forces, even in the simple case given, is tedious. It 
can readily be understood that in a complicated framework 
it would become intolerably so. Accordingly among archi- 
tects and engineers a method of greater simplicity is fol- 
lowed, leading to results practically close enough. This is 



120 



STATICS OF A RIGID BODY. 



done by considering the loads carried by the frame, includ- 
ing its weight, to be concentrated at the joints of the frame, 
and applying the conditions of equilibrium to each joint in 
succession. Thence the stresses along each piece, meeting 
at a joint, are found by the triangle of forces. 

The method belongs to a special branch of mechanics 
known as Graphical Statics. We add a short sketch of its 
application to the determination of the longitudinal stresses 
in simple trusses, forming as it does a good illustration of the 
triangle of forces. 

104. Jointed Frames. — As the simplest possible example 
Fiq | 09 of a jointed frame, let us consider 

three beams hinged by pins at A, B, 
0, and resting on supports at A, C in 
the same horizontal plane. This is 
known as a Triangular Truss. Sup- 
pose the beams all alike, and of 
weight W each. The reactions of 
the supports balance these weights 
and act vertically upwards, the sup- 
ports being horizontal. Hence the 
external forces acting and keeping the truss in equilibrium 
are as in Fig. 109. 

Next transfer the weights to the pins. Thus 



/- 




G l ^ 
G, 


\ 


' 










\ 


/ 


'W 


W 


,w 



JFat G, = i JFat A + JJPat B, 
W at G t = i W at B -f- \ W at C, 
IF at c7 3 = JJFat A + £TFat G\ 

. ' . sum = 

Wat A + IF at 2?+ TFat C; 
and each reaction being equal to 
half the total weight 3 W, we have 
the forces as in Fig. 110. 

Combining forces and reactions 
at A and G, we have finally the 
forces as in Fig. 111. 



Fig. 110 



-f-w 






L 


1 

1 




w \ 






IW 




fw 



JOINTED FRAMES. 



121 



We have thus transferred the weights of the beams to 
the joints, and can now consider the 
beams as without weight, and indi- 
cating direction only. The result- 
ing stresses in the pieces we next 
find. 

Since the weights on each pin are 
in equilibrium with the stresses pro- 
duced in the pieces meeting at the 
pin, we consider the pins one at a 
time. 

(a) Pin A. The forces acting are W/2 vertically up- 
wards, and the unknown stresses in AB, AC. Draw (Fig. 
112) Oa to scale to represent 1)72. From a draw ab parallel 
to BA and b parallel to A C. Then ab, b represent on the 

Fig. 112 






same scale the stresses in AB, AC. Their directions are 
indicated by the arrow-heads. 

(b) Pin B. The forces acting are the stress in A B and 
W which are known, and the stress in BC which is un- 
known. DrSLwba to represent the stress in AB, ac to repre- 
sent W; then cb will represent the stress in BC. Xotice 
that the arrows on ab point in opposite directions in Figs. A 
and B. This is because the stress on the pin A is opposite 
to that on the pin B. 



122 STATICS OF A RIGID BODY. 

(c) Pm C. The forces are the stress in BC, W/2, and 
stress in A C, all of which are known. For check the dia- 
gram may be drawn as in Fig. 0. 

105. It is evident that we should have saved labor by 
adding the second figure to the first and the third to the 
sum as in the fourth figure, which is the complete Stress 
Diagram. 

In practice it is convenient to consider the stress dia- 
gram as in two parts. Thus the line ac is the polygon 
of external forces, ac being the downward force at B bal- 
anced by the upward forces eO at C and Oa at A, and is 
complete in itself. The closing of this polygon shows that 
the reaction forces have been properly estimated. On this 
polygon as base the stress diagram is added step by step 
until complete. 

In case the truss is symmetrical, as in our example, it is 
only necessary to consider the first half of the pins. But 
it is safer to consider all of them, as the symmetry of the 
drawing will furnish a test of its accuracy. 

106. In Fig. A we see that in the beam AB the stress 
acts towards the pin A and also towards the pin B. The 
beam is thus compressed between the pins, and is called a 
Strut. Similarly for BC. In A C the stress is from the pin 
A and also from C, and the beam is in a state of tension. 
It is called a Tie, and in practice a rod would be used. 
Hence in designing a structure to carry an assigned load a 
study of the stress diagram will show not only the amount 
of stress but the kind of stress, and therefore whether a 
strut or tie should be employed. 

107. For tracing the connection between the pieces them- 
selves and the stresses in them as shown by the stress dia- 
gram, an exceedingly convenient system of notation has 
been devised.* 

* Due to Prof. Heiirici, Loudon, but usually known as Bow's 
notation. 



JOLSTED FRAMES. 



123 



A beam or a stress is named by letters placed on either 
side of it. Thus (Fig. Ill) Oft is the tie A C, Oa the reac- 
tion TT/2 at the left support, aft the strut AB, ac the load 
IT at B, and so on. These letters carried into the stress 
diagram (Fig. 112) give us aft the stress in the piece aft] eft 
the stress in eft, Oft the stress in the rod Oft. The letters 
A, B, C at the pins do not enter the stress diagram. 

108. In the following examples first draw the truss to 
scale (inches to the foot) from the dimensions given, next 
compute the pin loads, next the reactions of the supports, 
next draw the force polygon (pounds to the inch), and 
finally the stress diagram. Scale off the stresses, and tabu- 
late under the heads Compression and Tension. 

Ex. 1. In a triangular roof truss the rafters are 2^ ft. 
apart and the roofing material weighs 20 lbs. per sq. ft. 
The span is 24 ft. and height 5 ft. : find the stresses on the 
rafters. Ans. 590 pounds. 

2a. Show that the stress diagram for the truss repre- 

Fig. 113 




F ;•■■■;•.= :-.s 




sented in the figure, loaded at the centre over the vertical 
piece aft, known as the king-post, is as in the margin. 

2b. A foot-bridge (Fig. 113) 18 ft. span and 6 ft. breadth 
has a crowd of people on it equal to 100 lbs. per sq. ft. of 
floor surface. The king-posts of the two trusses are 3 ft. 
in depth : find the stresses. 

Ans. Stress on post aft — 2700 pounds. 

3. In (2a) the span is 21, depth d: show that the eompres- 



124 



STATICS OF A RIGID BODY. 



sion in Aa is Wl/2d, and find the tension in Oa and the 
stress in the vertical ab. 

4. In a roof of 30 ft. span and height 10 ft. the trusses 

Fig. 114 




Fig. 115 



are 10 ft. apart, and the pieces ab, cd come to the middle 
points of the rafters. If the weight of the roof covering 
is 25 lbs. per sq. ft. surface, draw the stress diagram and 
scale off the stresses. 

5. Draw the stress diagram 
for a Grennan truss loaded as 
in Fig. 115. 

Qa. Draw a stress diagram for 
a queen-post truss (Fig. 116). 
The queen-posts ab, be divide the 
span into three equal parts, and 
the truss is loaded at the joints 
with weights W. 

6b. Afoot-bridge (queen-post) 
of span 25 ft., breadth 7 ft., 
length of queen -posts 3 ft., car- 
ries a load of 150 lbs. per sq. ft. of floor: find the stresses 
developed, the queen-posts Fig. ii6 

dividing the span into 
three equal parts. 

7. Draw stress diagrams 
for the roof trusses repre- 
sented; the first being 
that of the Rock Island 
Arsenal and the second ° 

the Masonic Temple, Philadelphia (Figs. 117, 118). 





JOINTED FRAMES. 
Fig. 117 



125 




109. The graphical method may be conveniently applied 
to finding the stresses in a mechanism. Take, for example, 
the steam-engine. Let Fig Il9 

P be the pressure ex- 
erted by the piston on 
the pin A (Fig. 120) 
of the cross-head. It 
is transmitted by the 
connecting-rod to the 
crank-pin B, and 
thence to the crank axis 0. If now the machinery is driven 
by a wheel OG on the axis C working in another wheel at G, 
the resistance R would be tangent to the pitch circles of 
these wheels. [This includes the case of a locomotive when 
the traction is exerted by adhesion to the rail XY. The 
amount of this adhesion corresponds to 7?.] 
- Consider the pin A. It is in equilibrium under the 
pressure P along the axis of the piston-rod, the thrust Q 




126 



STATICS OF A RIGID BODY. 



along the connecting-rod, and the reaction JV of the guide- 
bar of the cross-head. Hence plot P to scale, and complete 
the triangle of forces, from which scale off Q and JV. 

Again, the wheel CG is in equilibrium under the action 
of Q, R and the reaction S of the crank axis C. All three 
must meet in the point D, where Q and R meet. Hence 
plot the triangle of forces, and scale off S and R. The 



Fig. 120 





relation between P the piston pressure and R the force 
transmitted to the mechanism is therefore determined. 

We have neglected the weights of the pieces. The only 
one important to consider is the weight resting on the 
crank axis C. Call it W. It acts vertically. Combine Q 
and W into one resultant R i . The reaction S will now pass 
through the intersection of i?, and R. Hence complete 
the triangle of forces for R, S, R , , and scale off S and R. 
Thus the relation between P and 7? is found. 

Similarly, the weights of all of the pieces may be taken 
into account if desired. 



ATTRACTION. 127 

110. Attraction. — In astronomy and mathematical elec- 
tricity an important application of the composition of 
forces is to find the resultant attractive force between a 
finite body and a particle, the law of force being that 
known as the "law of inverse squares." This law maybe 
stated as follows: Every particle- of matter in the universe 
attracts every other particle with a force ivhose direction is 
in the line joining them, and ichose magnitude varies di- 
rectly as the mass of each particle and inversely as the 
square of the distance between them. Thus if the masses 
are m, m x , and r. the distance between them, the mutual 
attraction F is given by 

F = c?n?n J /r i , 
when c is a constant. 

The attraction of one particle on another being an inde- 
pendent attraction, the total attraction of a number of par- 
ticles (or body) on a particle is a problem of summation 
which is most readily solved by means of the integral cal- 
culus. We shall develop one problem to illustrate the 
method. 

Ex. 1. To find the attraction of a uniform thin circular 
disc, radius a, thickness h, and density 6, on a particle 
of unit mass situated on a line OC through the centre of 
the disc and perpendicular to its plane. 

Let the distance 00= b. Conceive the disc divided into 
an indefinitely great number of concen- 
tric rings, centre C, and let x be the radius 
of any one of the rings and dx its width. 
Then mass of ring = Inxdxhd, and dis- 
tance of all points of the ring from = 
Vb' 2 -j- x 2 . Hence 

attr. of ring along OC— cX ^Ttdhxdx cos COD/{b 2 + x 2 ), 

and result, attr. of disc = '2 n debit J xdx/(b 2 -h 35 *) 

= 27cdch}l-(b/b* + f Jt )*\, 

an important result in mathematiciJ electricity. 




128 STATICS OF A EIGID BODY. 

2. If the disc is of indefinitely great extent, show that 
the attraction = "In Sell. 

3. If the particle is indefinitely near to the disc, show 
that the attraction = 2x6 ch. (See Thompson's Electricity 
and Magnetism.) 

4. Hence show that all parallel discs of the same thick- 
ness, forming the bases of cones having the same vertical 
angle, exert the same force on a particle at the vertex. 

5. Hence show from (-1) that the resultant attraction of 
a thin spherical shell of uniform density on a particle situ- 
ated within it is nil. 



CHAPTER V. 



FRICTION. 

111. As already stated, greater simplicity and clearness 
are secured by considering the properties of bodies one at a 
time, and thus leading up to the actual state of the case — 
which is quite complicated, since bodies in nature possess 
many properties. Thus far the surface of a body has been 
assumed to be perfectly smooth, that is (Art. 63), to exert 
a pressure in a normal direction only, or what is the same 
thing, to offer no resistance to the motion of a body pressing 
against it. Bat in reality we know that if one body be 
moved along another (as a book along a table) a certain 
resistance will be offered to the motion. The resistance 
arises from irregularities in the surfaces in contact, from 
elevations and depressions .which fit more or less closely 
into one another. To it the name Friction is given. 

Suppose a body of weight W to rest on a horizontal table, 
and to be pressed by a vertical force Q (as of a load resting 
upon it). The total pressure P (= Q -\- W) is balanced by the 
vertical reaction N of the p . [22 

table. If now a force is ap- 
plied parallel to the table, 
and gradually increased, a 
magnitude i^will be reached 
when the body is just on 
the point of moving. The 
body is held in equilibrium 
by the force P, the force F, 
and the reaction B, which 
can be no longer vertical. Let R be resolved into vertical 
and horizontal components. The vertical component must 

139 




130 FKICTIOX. 

be equal to P, and is therefore equal to the reaction N. 
The horizontal component / must be equal to the hori- 
zontal force F. This horizontal component arises from the 
frictional resistance, so that we may treat friction as if it 
were a force precisely similar to the acting force F, but of 
the contrary sense. The body being just on the point of 
moving, the friction resulting is named Static Friction. 

From experiments carried out with this form of apparatus 
it is found that between surfaces with little or no lubrication, 
under moderate loads, and just beginning to slide on one 
another, the amount of static friction is — 

(1) Proportional to the normal pressure between the stir- 
faces in contact. 

(2) Independent of the areas in contact. 

(3) Dependent on the material of which the bodies are 
composed. 

These are known as the laws of static friction. They are 
roughly true, not only when the motion is on the point of 
occurring, but also when the velocity of motion is small. 
Hence the friction / corresponding to the normal pressure 
i^may be found from 

f=»N, 
where fj. is a constant. It is called the Coefficient of Friction, 
and its value must be determined by experiment. 

112. The above " laws" of friction were deduced from 
experiments made with surfaces having little or no lubrica- 
tion, and moving with low velocities; and for such condi- 
tions only are they to be depended on. In machines, how- 
ever, surfaces without lubrication and moving with low 
velocities are the exception, and we there have an entirely 
different set of conditions. The friction is now Kinetic* 

* The distinction between static and kinetic friction was first 
pointed out by Coulomb (1736-1806). The laws of static friction 
were enunciated by Coulomb, and confirmed by the later experi- 
ments of Morin at Metz in 1837-1838. 



ROLLING FRICTION. 131 

Recent experiments show that even with surfaces of the 
same material, the character of the lubrication, the load, 
the velocity, the forui of the surfaces, whether flat or curved, 
the areas as in contact, and the temperature of the surfaces 
have each great influence on the friction produced. No 
general relation between the friction (/) and the pressure 
(N) producing it has yet been deduced depending on these 
conditions. Special experiments are necessary in all cases 
where the conditions differ in any of the points mentioned 
from those entering into experiments already made. 
We may, however, in general, write 

/ = vk 

when the value of /* is determined by the conditions of the 
problem. These conditions will show whether we may as- 
sume the so-called laws of statical friction or have recourse 
to special experiment. 

Roughly, the coefficient of friction pt for well-lulricated 
surfaces is from -J to T V that for dry surfaces ; and if the 
pressure and velocity are not very large, it varies inversely 
as the pressure, directly as the area in contact, and inversely 
as the temperature. 

Ex. In a locomotive the engineer applies the brakes suffi- 
ciently to prevent slipping of the drivers in order to obtain 
the maximum brake retardation. Explain. 

113. When one surface rolls on another, the resistance 
encountered is termed Rolling Friction. In mechanisms it 
is in general small, and need only be considered in very 
special cases. In a locomotive the rolling friction between 
the wheels and the rails, like the journal friction of the 
wheels, tends to diminish the adhesion of the engine on the 
rails. 

114. The following coefficients of friction may be re- 
garded as average values, to be used when no special experi- 



132 



FRICTION. 



Fig. 123 



ments covering the cases under consideration are possible. 
The circumstances may be such that the tabular values are 
very far from the truth in these cases. Indeed, at present 
we may be said to be acquainted with no quantitative laws 
of friction of much value. 

Metal on metal, 0.10 to 0.30 

Metal on wood, 0.10 to 0.60 

Wood on wood, 0.10 to 0.70 

Wood on stone, 0.30 to 0.60 

Stone on stone, 0.40 to 0.75 

115. In the experiment of Art. Ill, suppose the friction 
/ and the reaction JV combined into a 
single resultant R. Let denote the 
angle which R makes with the normal 
to the table. Draw to scale and com- 
plete the triangle of forces F, P, R. 
Then 

tan <f>=F/P=f/N=jji. 

Hence has always the same value so 
long as the substances in contact are the 
same and under the same conditions. 
It is called the Angle of Friction. 

The fact that the reaction force R, 
which holds the applied forces i^and P 
in equilibrium, makes with the normal 
to the surfaces in contact an angle equal to the angle of 
friction, gives a key to the application of the graphical 
method of the polygon of forces to problems involving fric- 
tion. The method is especially valuable in cases where the 
forces are interlaced as in mechanisms. 

Ex. 1. The weight on the driving-wheels of a locomotive 
is twenty tons, and the coefficient of friction is 0.2: find the 
greatest pull the engine is capable of. Ans. 4 tons. 

2. A train moving at 40 miles an hour is brought to rest 




ANGLE OP REPOSE. 



133 



by friction in half a minute. Prove that the coefficient of 
friction is 11/180. 

116. If the reaction iVis constant the direction of R will 
depend on the value of f, that is, of F. Hence, if F is such 
that the inclination of R to the normal exceeds the angle 
of friction, motion will take place. This gives us the clue 
to an experimental method of determining 0. 

Suppose a body of weight W to rest on a table, and 
that the table is tilted 
through a a angle 6 
about the edge A. We 
have now the bo'dy 
resting on an inclined 
plane. Continue tilt- 
ing until the body is 
just on the point of 
moving down the plane. 
At this point the forces 
holding it in equilibrium are W, N, and /, which act as 
shown by the arrows. DraAv the triangle of forces, and 



Fig. 124 




But 



f/lt=tsmff. 

//iV r =tan (p. 



Hence, 6 = <p, or the angle of inclination of the plane, is 
equal to <p, the angle of friction. For this reason the angle 
of friction is called the Angle of Repose. 

Ex. 1. Deduce the relation between 6 and by resolv- 
ing the forces vertically and horizontally. 

2. A boy's sled weighs 10 lbs. To pull it over a horizontal 
stretch of snow requires a force of 1 pound : find the eoeff. 
of friction. Ans. 0.1. 

3. Is it correct to say that between smooth surfaces the 
-pressure is normal, but when friction occurs the pressure 
is inclined to the normal at the angle of friction? 

[No. Pressure between surfaces is ahrays normal. The 



131 



resultant of the pressure and friction makes an angle with 
the normal.] 

4. To find the angle (3 which a given force F must make 
with the normal that a body of weight W may just be on 
the point of sliding on a horizontal surface. 

Ans. Fsin (/i-f 0) = IF sin 0. 

5. In Ex. 4 find for what value of j3 the force F is the 
least possible. Ans. (3 — 90° — 0. 

6. Find the force necessary to drag a body of weight W 



Fig. 125 




up the incline A C, the 
inclination 6 of the 
plane and the direction 
(3 of the force being 
given. 

[The force / will act 
down the plane. The 
forces acting are F, 
W, R. Form triangle 
of forces, and scale off F and R. Or compute from 
F cos (f3 — <p) = W sin (6 -f 0). Get the same answer by 
resolution of forces.] 

7. If the force is horizontal, prove 

F = IF tan (# + 0). 

8. If the force is parallel to the plane, prove 

F= TF(sinfl ± fx cos 6) 

as the body is on the point of moving up or down the 
plane. 

9. In Ex. 6 find the force necessary to push the weight 
down the plane. Ans. P cos ((3 -\- 0) = W mi (6 — 0). 

10. The force necessary to haul a train at uniform speed 
on a 1% grade is 3.5 times that on the level. Show that the 
coefficient of friction is 1/250. 

11. The angle of a wooden incline is 68°. Show that it is 
impossible to drag a wooden block up the plane by a hori- 
zontal force, the coefficient of friction being 0.4. 

12. The foot of a ladder of length / rests on the ground 
at A and the top at B against a rough vertical wall : find 
its inclination 6 when on the point of sliding, the coeffi- 
cient of friction in each case being 0.5. 

Ans. tan 6 = f-. 



MOTIOX OX A ROUGH SURFACE. 



135 



117. Motion on a Rough Surface. — Suppose a particle 
of mass m to .slide on a rough horizontal plane of which 
the coefficient of friction is }x ; then the normal pressure 
beiDg mg and the friction pwig, we have for the accelera- 
tion of motion 

a — — jjmg/m = — jug. 

\fu is the initial velocity, we have, by substituting in the 
equations of Art. 13, 

v =u - jugt, ■ 

v 1 = u 2 — 2jugs, 
s — ut — \jxgtf, 

and the motion is completely determined. 

If the plane is inclined at an angle 6, then resolving mg 
into its components mg sin 6 along the plane, and mg cos 



Fig. 126 




at right angles to it, the latter repre- 
sents the pressure on the plane. Hence 
the friction is jxmg cos 6, and the 
force causing motion down the plane 
is mg sin — jxmg cos 6. Therefore 
the acceleration down the plane is 
found from 

a = (mg sin — fimg cos 6)/m 
= g (sin 6 — jx cos 6) 
= g sin (0 — 0)/ cos 0, 
if is the limiting angle of friction. 

Ex. 1. A body with initial velocity u slides along a rough 
horizontal plane on which the coefficient of friction is >u: 
show that it will come to rest in ii/j.tg seconds after pass- 
ing over a distance ir/'liui. 

2. If a body is projected up a plane whose inclination is 
6, show that a — — g sin (6 + 0)/ cos 0. 

3. A train of w lbs. is hauled along a horizontal track by 
a constant pull of p pounds. If the resistance of friction 



136 



is/ pounds, find the velocity of the train in t seconds after 
starting from rest, and the distance passed over in that 
time. Ans. s = %(p —f)gt*/w. 

4. A train of 100 tons (excluding engine) runs up a 1% 
grade with an acceleration of 1 ft. per sec. If the friction 
is 10 pounds per ton, find the pull on the drawbar between 
engine and train. Ans. 4| tons. 

We now pass to various practical applications of the 
principles laid down. 

Fig. 127 




118. Axle, Journal, or Pin Friction. — A beam AB is 
pierced by a shaft 0, about which as an axis it is acted on 
by two forces P, Q, of which P is the driving force and Q 
the resistance. 



AXLE, JOURNAL, OR PIK FRICTIOX. 137 

Suppose P to be on the point of overcoming Q. The 
pressure N between the two surfaces being normal, passes 
through the centre 0. The resultant R of JVand the fric- 
tion / must pass through C, the intersection of P and Q, 
and at the same time be inclined to the direction of J\ 7 at 
an angle 0. Hence plot P to scale, construct the triangle 
of forces, and scale off Q and R. The point A, through 
which the resultant R passes, is the point where the 
pressures on the axles are concentrated. The reaction N 
and friction f at A may be found by completing the tri- 
angle of forces to the left. 

Notice that if we drop a perpendicular OB on R and 
describe a circle with radius OB, the direction of 7? is a 
tangent to this circle. But OB = r sin (p, a known quan- 
tity. Hence, to plot R describe a circle of radius = r sin cp 
— fir nearly, and a tangent from C, the intersection of P 
and Q, to this circle will give the direction of R. This circle 
is known as the Friction Circle.* 

From C two tangents may be drawn to the friction circle. 
The one to be chosen can always be found by considering 
the direction of the driving force and the consequent direc- 
tion of the friction (and therefore of R), between the jour- 
nal and bearing to oppose it. 

If the axle rotates uniformly, the solution is the same. 
For when it is just on the point of motion the condition of 
equilibrium is that the moment of the driving force is 
equal to that of the resistance. If now the axle revolves 
uniformly, the force acting must be a centripetal force, 
and the moment of this force about the axis being nil, the 
condition of equilibrium remains unaltered. 

Ex. 1. If P and Q are parallel, show that 

f=M{P+Q), nearly- 

2. Show that the pull on the drawbar of a train 
due to journal friction is per ton weight of the train 

* The friction circle was first given by Rankine (1820-1872). 



138 



FRICTIOX. 



= 2000 /ir/r t pounds, when i\ is the radius of the journal, 
i\ the radius of the wheel, and /* the coefficient of friction. 

119. A good illustration of sliding and axle friction in a 
mechanism is afforded by the reciprocating parts of a steam- 
engine. (See Figs. 119, 123.) 




Draw the friction circles at the pins A, B, C. The thrust 
Q along the connecting-rod lies in the common tangent to 
the friction circles at A and B (Art. 118). 

Consider the pin A. It is in equilibrium under the 
piston pressure P along the axis of the piston-rod, the 
thrust Q of the connecting-rod, the sliding friction f, and 
the reaction A^at the cross-head guides. The resultant #, 
of /and N makes the angle of friction <p with the normal 
to the sliding surface. Its direction must pass through 0, 
the intersection of P and Q. Hence plot P to scale, com- 
plete the triangle of forces, and scale off Q and R x . 

Again: the wheel CG is in equilibrium under Q, R, and 
the reaction S of the crank axis C. All three must meet 
in the point O x , where Q audi? meet. Hence plot the tri- 
angle of forces for Q, S, R, and scale off the values of S 
and R. 



BELT FRICTION. 



131 



The relation between P and R is therefore determined. 
Ex. Draw the stress diagram if the weight W of the 
wheel CG is taken into account. 

120. We have already found the ratio of R to P when 
friction was neglected (Art. 109). This was a purely 
theoretical value. The actual case in which friction oc- 
curs is that just considered. The ratio of R to P is here 
less than in the former case. The ratio of the latter or 
actual value to the theoretical value is known as the effi- 
ciency of the mechanism. (See Art. 139.) 

121. Belt Friction. — Consider a belt or rope passing over 
(or around) a cylindrical block securely 
fastened, and let P 1} Q l denote the forces 
at the ends of the belt. Suppose P 1 to be 
just on the point of overcoming Q x , or 
that the belt is moving with uniform 
velocity.. The forces acting on the belt 
are P 1} Q l3 the reaction of the cylinder 
and the friction of the cylinder. In 
order to find the relation between them, 

conceive the arc of contact 6 of the belt cut into elements 
of indefinitely small 



Fig. 129 




length els, find the rela- 
tion for one element, 
and then sum up for 
the whole arc. Con- 
sider now a single ele- 
ment ah of the belt, and 
let P, P + dP be the 
tensions at the points 
a, b ; p the normal pres- 
sure per unit of length 
of arc, and therefore 
pds the pressure on ab, whicl 
and jipds the i'rictional r 



Fig. 130 




may be taken to act at a 
istance on ab. Call a Ob = d6. 



140 



Then, neglecting the weight of the belt, 
ppds = (P + dP) cos dd- P = dP, ultimately; 
pds = (P + <^P) sm dd — Pddj since <:/# is small. 
Eliminating pds, we have 

dP = /«/W. 
Summing up all of the elements of the rope in contact 

with the cylinder, 

pp. pe 



log. 



M 



^ 



or P x 

when e is the base of the natural system of logarithms 
(= 2.718) and 6 is the angle subtended by the arc of con- 
tact between the rope and the cylinder expressed in circular 
measure. 

Transferring to the common system of logarithms, as 
more convenient for computation, 

log.. *\/«i = 0.434M 

If the weight Q t is on the point of slipping down instead 
of being drawn up, the action is the same as if P x and Q 
changed places, and therefore Q x = P^er. 

The conditions are the same and the relation between 
P, and Q x is the same if the cylinder instead of being fixed 
is capable of turning about 0, and no slipping occurs, axle 
friction being neglected. 

122. Suppose, for example, the pulley B drives the pulley 
Fig. 131 A of radius r by means of 

a belt passing round them. 
Let P, Q denote the ten- 
\ sions of the belt on the 
' two sides of A. The ef- 
fective driving force is 
q P — Q, the difference of 

the tensions. This may 




BELT FRICTIOX. 141 

be balanced by the raising of a weight W on a concentric 
pulley of radius s. Just at the point of motion we have 
Ws = (P- Q)r, 

from which equation, along with P = Qe 1 * 9 , P and Q may 
be found. (See Art. 138.) 

Ex. 1. A rope passing over a wooden cylinder supports 
a barrel of flour weighing 196 lbs., find the force which will 
just raise the barrel, the coefficient of friction being 0.4. 

Ans. 689 pounds. 

2. In Ex. 1 find the force that will just keep the barrel 
from slipping down. Ans. 56 lbs. 

3. Find the number n of turns of rope round a snubbing- 
post that a man pulling P pounds may just be able to hold 
a canal-boat pulling Q pounds. Ans. 27tjun = log e P/Q. 

4. A chain is wrapped twice round an iron drum : find the 
coefficient of friction if a pull of 100 pounds just supports 
50 tons. Ans. 0.55. 

5. If jj. = 0.25, and a rope passes twice round a post, prove 
that any force will balance another more than twenty times 
as great. 

6. In the St. Louis cable road the contact is two wraps of 
the cable on the driving drum, which is 10 ft. diameter: 
compare the tension of the taut side of the cable with the 
tension of the slack side. 

123. Effect of Centrifugal Force. — The relation between 
the tension and pressure at any point results from the 
second equation on p. 140. For 

P = pds/dd = pr. 

When a belt is run at a high rate of speed, the effect of the 
centrifugal force of the belt is to lessen the normal pressure 
on the pulley. If w is the weight of 1 unit length of belt, 
and v the velocity, the centrifugal force per unit length is 
wv^/gr where r is the radius of the pulley. Hence the nor- 
mal pressure^ per unit on the pulley is P/r — wv*/gr, and 

P =pr + wv*/g, 



142 




or the tension is increased by wv*/g. Hence the greatest 
tension instead of being P x is P,+ wv ' Vff> wnen centrifugal 
force is taken into account. 

This value, substituted for the greatest tension in the pre- 
ceding formulas, will enable us to find the belt dimensions. 

124. Pulley Tackle. — We may discuss the friction of 
pulley tackle as a special case of axle fric- 
tion. The resistance arises mainly from 
the friction of the axles of the sheaves. 

Consider the simple case of the fixed 
pulley. Let F be the driving force, 
W the weight of the body to be raised, 
r radius of pulley axle and a radius of 
pulley sheaf. 

Describe the friction circle of radius 
j.ir and centre 0. The resultant pressure 
^^ is tangent to this circle and parallel to F 

and IF. If F is on the point of moving down, its position 
is as in the figure. Take moments about A, 

F{a - fir) = W(a + M r), 

or F — W(a + /*>')/(« — M r ) 

= (1 -}- 2/ir/a) W, nearly, 

= X W } suppose, 

where the coefficient A is introduced for convenience of 
writing. It gives the relation between F and W. 

Ex. 1. If the axle were smooth, then F— W. 

2. The efficiency = 1/A. 

3. The effect of the friction is the same as the raising of 
an additional weight (A — 1) IT if the axle were smooth. 

4. Find the tension of the cord supporting the pulley. 



DIFFERENTIAL PULLEY. 



143 




In the derrick represented in Fig. 133, where the rope 
passes under the movable pulley and over 
the fixed, we have the tensions as follows: 
Fthe driver, F/X the first driver, F/X 2 the 
second driver. Hence, considering the cords 
parallel, 

W=F(\+1)/X\ 

Ex. 1 . If friction is neglected, show that 
W=2F. 

2. The efficiency = (A + 1)/2A 2 . _ 

3. Solve the problem by a graphical con- 
struction. 

4. With two double sheave blocks 

W= F(X b - 1)/(A B -A 4 ). 

5. With two double sheave blocks, neglecting friction, 
W = 5F. Hence find the loss due to friction. 

6. In a double sheave tackle the under block is fixed and 
the upper movable : find the pull on the support A, and the 
loss due to friction. 

125. The Differential Pulley consists in the upper block 
of two . sheaves, radii a, b, fastened together, and in the 
lower block of a single sheaf of diameter a -\-b. An endless 
chain passes round the sheaves as in the figure. Notches 
are cut or teeth set in the upper block sheaves which fit 
the links of the chain. 

Let F be the force applied, and IT the weight which is 
on the point of being raised. Consider the lower pulley. 
One chain is driver the other is driven. If F x denotes the 
tension of the driver, the tension of the driven is XF X , from 
Art. 124. Hence 

W=F l + XF 1 (1) 

On the upper block F and F t drive and A/ 7 , is driven ; 
J 7 and XF t acton the larger sheaf, and F y when reduced to 
this sheaf is FJb/a. Hence 

F+ FJj/a = A x XF = X 2 F } . 



144 



FRIOTION. 




Fig. 134 




Substitute for F t its value from (1) and 

F=W(Va-l)/a(l + X), 

which gives the relation between F and W. 

If the driving force F is removed, that is, if F = 0, then, 
whatever be the weight, we must have for equilibrium 

\*a - b ' = 0, 

which gives the relation between the radii, that this may be 
possible. If \ — 1.1 then b/a = 1.2, and the diameters of 
the pulleys are roughly as G to 5. For these dimensions 
the chain will not slip, whatever weight is being raised. 
The practical value of the pulley lies largely in this circum- 
stance. 



THE SCREW. 



145 



Ex. 1. Examine a pulley of this kind in a machine-shop. 

2. In a differential pulley the diameters of the pulleys 
of the compound sheave are a, b in. : find how many revo- 
lutions are required to raise the weight c inches. 

Ans. 2c/7t(u -f b). 

126. The Screw. — In the cases of axle friction considered, 
the motion has been simply a motion of rotation, — either the 
bearing rotating about the axis as in the pulley, or the axle 
rotating in the bearing as in shafting. We may conceive, 
however, the axle not only to revolve, but to advance in the 
bearing, or the bearing to revolve and advance along the 
axis. The combination of the two motions gives rise to 
Screw Motion. 

Thus suppose H to be any point on the circumference of 
a cylinder AB which revolves in a bearing, or nut and 
at the same time advances 
uniformly along the axis. 
AB. By the motion of 
rotation alone, while the 
cylinder makes a revolu- 
tion, H would describe a 
path equal to the circum- 
ference. Develop this in 
the line HE. But the mo- 
tion of translation carries 
it a distance equal to EF. 
Hence it is found at K, 
and the path HK would be traced by wrapping the tri- 
angle HEF about the cylinder. The path of the point II 
is called the Thread of the screw, and the distance HK or 
EF between consecutive threads the Pitch. 

The inclination of the thread to the axis is given by the 
angle EHF(= fi). JS T ow 

tan (3 = EF/HE 

b= pitch/circum. of cylr., 




146 



a constant quantity. Hence the inclination of the thread 
to the axis is always the same. 

Consider a screw-jack with a force F applied at the end 
of a lever I, and that a weight W is on the -point of being 
raised. 

The force F with lever arm I is equivalent to Fl/r with 
lever arm r, or acting along the thread. The screw is in 




equilibrium under the action of W vertical, Fl/r horizon- 
tal, the reaction N inclined at the pitch angle /3 to the axis, 
and the friction /^Y along the thread. The resultant R of 
jYand f-iN makes an angle with the normal. Hence lay 
off W to scale, complete the triangle of forces, and scale 
off Fl/r and JR. Or from the triangle we have at once 



Fl 
r 



/ir=tan(/?+0), 
Fl = Wr tan (/3 + 0). 



THE SCKEW. 147 

Without friction, equilibrium exists between Fl/r, N 9 and 
W. Hence abcl would represent the triangle of forces, and 

efficiency = g = |/g = tan /3/tan (ft + 0) 

Ex. 1. Deduce the relation between F and W by resolu- 
tion of forces. 

2. If /3 = 0, the efficiency is null. Explain. 

3. In what other case is the efficiency null ? 

4. Show that the efficiency of a screw is greatest when 
the pitch angle is 45°, nearly. 

5. In a screw the pitch angle is 45° and the coefficient of 
friction 0.16 : find the efficiency. Ans. 0.72. 

6. If tan /3 = 0.1, tan = 6.01, prove efficiency = 0.9; 

tan/? = 0.1, tan = 0.1, " " =0.5; 

tan /? = 0.1, tan = 0.2, " " =0.3. 

Hence show the importance of lubrication. 



CHAPTER VI. 
WORK AND ENERGY. 

127. When an agent exerts force on a body the effects 
produced are change of position and change of form. The 
first of these has been considered in the preceding chapters, 
the effect of force being measured by the acceleration pro- 
duced. All of the results found depend on the experiment- 
al principles assumed, the laws of inertia, mass-accelera- 
tion, and stress. 

We now proceed to study the effects of force from another 
point of view, and to lay down a method of treating these 
effects more general than that stated hitherto, in that it 
applies to change of form as well as to change of position. 
The comparison of results reached by the two methods 
affords a test of the truth of the new method just as already 
pointed out, that the truth of the laws of motion is not 
capable of direct demonstration, but must be tested by con- 
sequences arising from assuming their truth. 

In order that change of position or change of form may 
occur by the action of a force, the body must yield to the 
force, or, in other words, the point of application of the 
force must be displaced. When this displacement occurs, 
Work is said to be done by the force and on the resistance 
offered. Thus when a body is falling freely, work is done 
by the force of gravity on the body; while a spring is being 
bent, work is done by the acting force on the spring; and 
so on. If a body is lifted vertically upwards, work is being 
done by the lifting force and against the force of gravity. 

148 



MEASURE OE WORK. 149 

If, then , we take work in the direction of a force as -\-, 
that done against a force must be taken as — . 

In lifting a weight W (that is, in overcoming the gravity 
force of W) through a height Ji, the work done depends on 
the values of W and h, and is measured by the product 
Wlu In general, the work K of a force F is measured by 
the product of the force and the displacement s of its point 
of application in the direction of the force, or 

K=Fs. 

This is the case whether the displacement is actually 
made or is only conceived to be made. In the latter case 
the work is said to be virtual (= hypothetical). 

128. In the definition nothing is said about the path of 
the point of application of the force. 
If, then, the point of application A of a 
force F acting along the line OA be dis- 
placed to B in any path, and BG be let 
fall perpendicular to OA, the distance 
AB is the total displacement, and- the 
distance AC is the displacement s of A 
in the direction of the force. Hence by the definition 

K= Fx AC 
= F X AB cos 
= Fcos d X AB-, 

or, the work done by a force acting obliquely to the path of 
a body is measured by the product of the force and the 
projection on its direction of the total displacement, or by 
the product of the component of the force along the total dis- 
placement by that displacement. 

When = 90°, then cos 6 = and K = 0. Hence when 
-the displacement is at right angles to the direction of the 
force, the work of the force is n il. 




150 WORK AXD ENERGY. 

Ex. In a pendulum find the work done by the pull of 
the rod on the bob as it swings to and fro. Am, nil. 

129. In the general formula for work 

K=Fs; 

taking F = 1, 8 = 1, we have K — 1; and therefore the 
unit of tvorh is taken to be the work done by unit force 
acting through unit distance. Two forms are in common 
use — the scientific or laboratory unit, and the engineering 
unit or unit of every-day life. 

The scientific unit of work in the F. P. S. system is the 
work done by a force of one poundal acting through a dis- 
tance of one foot, and is called the Foot-poundal ; in the 
C. G. S. system it is the work done by a force of one dyne 
acting through one centimeter, and is called the Erg. Thus 
a force of 10 poundals acting through 2 ft. will do a work 
of 20 ft. -poundals, and 10 dynes through 1 meter a work 
of 1000 ergs. To avoid inconveniently large numbers, an 
enlarged unit, the Joule, equal to 10 7 ergs, is often used in 
the O. G. S. system. 

The engineering unit of work is based on the gravitation 
measure of force, and is the work done by a force of one 
pound acting through a distance of one foot. This is 
usually called a Foot-pound. Thus the work done in raising 
100 lbs. vertically through 6 feet is the work done in over- 
coming a weight (= gravity force) of 100 pounds through 
6 feet, and is 600 foot-pounds. In the metric system the 
engineering unit is the kilogrammeter (kgm.), which is the 
work done by a force of one kilogram acting through a dis- 
tance of one meter. 

130. The dimensions of the unit of work will be the 
dimensions of F X s. The dimensions of F are ML/T 2 
(Art.ll) and of s, L. Hence the dimensions of iTare ML /T*. 



CONDITION OF EQUILIBRIUM. 



151 



This is the same result as that obtained for the dimensions 
of the moment of a force * (p. 95). 

Ex. 1. Prove 1 kilogr ammeter = 7.23 foot-pounds. 

2. Prove 1 foot-pound = 13.56 X 10 6 ergs, approx. 

3. Prove 1 foot-poundal = 0.42 X 10 6 ergs, nearly. 

" 4. The tractive force of a consolidation engine is 10 tons. 
Find the work done in hauling a train one mile. 

Ans. 105600000 ft. -pounds. 

131. When the direction of displacement of the point of 
application is not in the line of action of the 
force considered, the body must be acted on 
by other forces, and the work done estimated 
for each separately. Thus consider a particle 
of weight W to rest against a smooth vertical 
wall, and to be raised vertically by a force F 
acting at an angle 6 to the vertical. Let 
AB be the displacement. The work done 
against gravity is W X AB. The work done 
by the reaction i\ T is nil. The work done by 
i^isi^x BO. 

If the motion is uniform, that is, if the 
forces acting are in equilibrium, we have, by resolving 
vertically, 

F cos 6 - W = 0. 

Multiply each member by AB, and 

F cos 6 X AB - W X AB = 0, 

or Fx BO - WX AB = 0; 

or, when equilibrium exists, the sum of the works of the 
forces acting at the point is zero. 

* There is really a difference between the dimensions of the two, as 
unit work is the product of a force into a displacement in the direc- 
tion of the force, and unit moment is the product of a force into a 
distance perpendicular to the direction of the force, so that the 
latter is strictly y~—\MU/T** 





152 WORK AXD ENERGY. 

132. More generally, let several forces F x , F^> . . . act 
at a point causing a displacement 
OA. Let R be the resultant of 
the forces. 
h l S\ a From A let fall perpendiculars 

on the directions of the forces, and 

"% "^ , let l9 2 , . . . be the inclinations 

A of these directions to OA, and 6 

the inclination of the resultant R. Then (Art. 50) 

R cos 6 = F x cos X + F 9 cos % + . . . , 

or i2 X Oa/OA =F Y X Oh/OA + F, X O^/O^ + . . . , 

or J2 X Oa i = F x X Ob + F, X 00 + . . . ; 

or the work done by the forces is equal to that done by 
their resultant. 

The equation may be written 

= - R X Oa + F t X Ob + F n X Oc + . . . , 

which shows that the algebraic sum of the works done by a 
system of forces acting at a point and in equilibrium is 
equal to zero. 

Conversely, if any number of forces act at a point, the 
condition of equilibrium is that the sum of the works clone 
for every displacement shall be equal to zero. For the sum 
of the works is equal to the work of the resultant, and for 
equilibrium the work done by the resultant must be equal 
to zero for each and every displacement. This is the Princi- 
ple of Work as applied to forces acting at a point, or to 
forces acting on a particle. 



PRINCIPLE OF WORK. 153 

133. The principle of work may be regarded as included 
in the law of stress (Art. 27), the work done by the forces 
F l9 F 2 , ... corresponding to the action, and the work 
done by — R, or rather the work done against R, to the 
reaction. In this sense we have merely the law in another 
form, the action of the agent being measured by the work 
done by it, and the reaction of the resistance by the work 
done against it. Looked at in this light, the principle of 
work falls within our old lines, and is therefore consistent 
with them. 

134. It follows by summing up from particle to particle of 
a rigid body, that the condition of equilibrium is that the 
sum of the works done by the forces acting is for every dis- 
placement equal to zero. For, the body being rigid, no 
work is done as the forces are transferred from particle to 
particle, since there is no yielding. Hence it is necessary 
to take into account the external forces only. 

The same principle will -evidently apply to a system of 
bodies rigidly connected,or so connected that the geometrical 
relations existing among the parts are not disturbed by the 
displacement, and hence to what we call a Machine. A 
machine is so constructed that the configuration of the parts 
is not disturbed when it is in operation. 

The external forces, that is, the driving force and the 
resisting force, form a system 
in equilibrium; and hence the F ' 9- ,4 ° 

relation between them follows ^b 

at once by equating to zero 
the sum of the works done. 
The work done by the driving 
and resisting forces being con- 
trary in character, will be de- 
noted by opposite signs. | F 

Flx. 1. In the straight lever 
to find the relation between i^and W. 



W>f 



3 



154 



WORK and energy. 



[Let the point of application A of the force F descend a 
distance x, and B consequently ascend a distance y. Then 

Fx - Wy = 0. 

But from similar triangles 

x:y = AC:BC. 

.-. Fx A0=WX BC, 

which is the "principle of the lever" (Art. 88).] 

2. In a pulley tackle the driving force descends 1 ft., 
while the Aveight to be raised ascends 1 in. What force will 
raise 1 ton? Ans. 166§ pounds. 

3. In a bell-bottom jack-screw (Fig. 136) with force F 
applied at the end of a lever arm I a body of weight W is 
being raised with uniform velocity. Prove 

F X 2nl = W X pitch of screw. 

[Notice that while the lever arm makes one turn, the 
weight is raised a distance equal to the pitch of the screw.] 

4. Find the relation between F and W in the copying- 
press (Fig. 72), 21 being the length of the handle. 

Ans. F= W X pitch of screw/4^-/. 

5. In a telescopic jack-screw a smaller screw works in 
a companion nut cut in the 
larger screw B, which latter 
works in a nut in the fixed 
block B. The block A being 
fixed, the upper screw does 
not rotate. If I is the length 
of the lever arm, find the re- 
lation between F and W. 
Ans. Fx2ttI=zW.X din 3 , of 

pitch of screws. 
6. In a hoisting machine 
(Fig. 142) the gears are 36 to 
36 teeth, the drum 21 in. dia- 
meter, and the load for one 
horse 1J ton. Find the pull 
^ exerted by the horse at the end 
IP* of a 7 ft. horizontal lever. 
Ans. 375 pounds. 



Fig. 141 




>" 



WORK AGAINST FRICTION. 155 

7. In a derrick winch (Fig. SO) the crank is I in. leverage, 

Fig. 142 




Find the 



Fig. 143 



the gears n to 1, and the drum d in. diameter, 
two-man-power capacity, each man 
exerting a force of p pounds. 

Ans. 4,pln/d pounds. 
8. In a combination of single- 
threaded worm and wheel used in 
hoists, the worm wheel has n teeth, 
the lever handle is I in. long, and 
the radius of the drum around 
which the lifting rope winds is r 
in. Find the relation between F 
and W. Arts. Fin — Wr. 

135. To overcome friction, as to 
overcome any resistance, a certain 
driving force or effort is necessary. 
In a machine, therefore, work must 
be done to balance the friction that 
arises. Part of the effort commu- 
nicated is taken up in doing this work, which is, as it were, 
absorbed. This work may serve no useful purpose, and is 
therefore said to be wasted. This does not mean that the 
jvork is lost, as we shall see presently. 

The work absorbed by friction is measured as the work 




15G 



WORK AXD ENERGY. 



done by any other resistance. We have 

Work against fric. = fric. X dist. desc. == Pyus, 

where P is the normal pressure on the journal, p. the coef- 
ficient of friction, and s the distance described. 

Thus the work done in overcoming the frictional resist- 
ance in one revolution of a journal of diameter d and carry- 
ing a load of W lbs. is W/j. X red ft. -pounds. In case the 
journal revolves n times per minute the work absorbed is 
Wja X 7td X n ft.-pounds per minute. 

Ex. 1. Find the work done in hauling a sled weighing 
500 lbs. half a mile, the coefficient of friction being 0.2. 

Ans. 264,000 ft.-pounds. 

2. Show that the work done in hauling a body of weight 
W up a rough incline A O is equal 

to that done in hauling it along 
the level AB, the coefficient of 
friction being the same, and in 
raising it through the height BO. 
[For IF is equivalent to IF sin 6 
along the plane, and W cos 6 at 
right angles to it. Hence the 
force of friction along the plane 
is ^JFcos 6. The total effective 
force required to move the body up the plane is 

W sin + /i W cos 6. 

Work done in moving from A to O 

= (TFsin0-f fiiWco&0)AC 

= TFX BC+mWxAB, 

which proves the proposition. 

For moderate inclines the work done may be taken equal 
to WX BC+ptWxAC.] 

3. Find the work done in hauling a train of 100 tons one 
mile up a 1$ grade, the resistance being 8 pounds per ton. 

Ans. 2800 X 5280 ft.-pounds, nearly. 




WOKK OF A VARIABLE FOECE. 



157 



Fig. 145 



136. We have considered the force acting and the resist- 
ance to be overcome as uniform in their action. It more 
frequently happens that they are variable. For example, 
the pressure of a compressed spring, the pressure on the 
piston of a steam-engine before and after the steam is cut 
off, etc. In cases of varying resistance, the work done is 
the same as would be done by a force acting uniformly, and 
which is equal to the average of the varying forces. This 
average as in the case of velocities is most conveniently 
estimated graphically (Art. 8). 

Thus let AL j)lotted to scale represent the distance passed 
over by the point of application of 
the force, and let Aa plotted to scale 
represent the force at ^4. If the 
force is uniform throughout, the 
work done (= AL X A a) is repre- 
sented by the area of the rectangle A I. 

If the force is variable, let the distance A L be divided 
6 into a large number, n, of 

equal parts AB, BO, . . . and 
let Aa, Bb, . . . represent the 
corresponding forces. The 
average force acting through 
the distance AB may be taken 
to be \(Aa + Bb), through 
BO to be \(Bb + Oc), etc. Hence, by addition, 

Work done = - —(Aa -f %Bl -f 20c -f . . . + LI). 

If AB, BO, . . . are indefinitely small, the curve abc . . . 1 
becomes continuous, and represents the varying action of 
the force. It is called the Curve of Resistance. The total 
work done would be represented by the area AalL. 
~~ By means of certain contrivances the curve of resistance 
may be plotted mechanically by the resistance itself, as, for 




158 WORK AND ENERGY. 

example, in the steam-engine by the Indicator. Having the 
curve, the mean resistance A may be found by stretching 
a string so as to have equal areas above and below it. Or 
the area may be read off at once by an Amsler polar planirn- 
eter, and the work done found directly. Or the indicator 
drawing or "card" may be divided up by drawing equi- 
distant ordinates, the lengths of these ordinates scaled off, 
and the formula above applied. All of these methods are 
at times useful. 

137. Power. — It is important to notice that the term 
work, as defined above, is not in all respects the same as 
what is called work in ordinary language. Ordinarily the 
idea of time enters, and the idea of motion is not essential. 
A man merely supporting a load does not come under the 
mechanical definition, no matter how long he may support 
it, though he is doing work in the ordinary sense of the 
term. 

The definition is somewhat arbitrary, which is quite allow- 
able so long as we use the term consistently with the defini- 
tion. It is plain from the definition that any small force 
can do work of any magnitude provided sufficient time is 
given. Hence, in order to compare agents which do work, 
it is necessary to take the time employed into consideration. 

To indicate the amount of work performed in a given 
time, the term Power is used. By the unit of power we 
mean the power of an agent which can do unit work in one 
second. This is expressed as foot-pounclals per second, 
or ergs per second, or foot-pounds per second, according to 
the system of units employed. Thus to raise 1 ton of coal 
through 100 feet in 10 min. would require an expenditure 
of 2000 X 100 ft. -pounds of work in 10 min., or of 20,000 
ft.-pounds per minute, which would be the expression of 
the power of the agent. 

When the power of an agent doing work is great, no very 
definite idea is conveyed by the large numbers of the above 



POWER. 159 

units that indicate this, and accordingly multiples of 
the units are used. In engineering work the multiple 
unit employed is called a horse-power. The term horse- 
power was introduced by James Watt. As horses formerly 
did the work done by steam-engines, it was natural to 
institute a comparison. Watt found that a Clydesdale 
horse could walk 2| miles an hour, and at the same 
time raise a weight of 150 lbs. This is equivalent to 
%\ X 5280 X 150/60 = 33000 foot-pounds per minute. 
The agent, therefore, which could do a work of 33000 ft.- 
pounds per minute, or 550 ft. -pounds per second, was 
named a Jwrse-power. 

The enlarged unit used in electrical work is the watt 
It is the power of an agent which can do a work of one 
joule (= 10 7 ergs) per second. 

The relations between these units will be seen in the ex- 
amples. 

Ex. 1. Show that one H. P. == 550 ft. -pounds per sec. 
= 396,000 inch-pounds per min. 
= 198 inch-tons per min. 

2. Show that one H. P. = 5o0g ft.-poundals. per sec. 

3. Show that one H. P. = 746 watts. 

4. Show that one joule = f ft. -pound, approx. 

5. The French H. P. is 75 kilogrammeters per sec; show 
that it is about -fa less than the British H. P., and that it is 
equal to 735} watts. 

6. Show that one watt = f ft.-pound per sec, approx. 

7. 22 tons of coal are to be hoisted through 50 yards in 
10 min.: find the H. P. of engine necessary. Ans. 20 II. P. 

8. How many gallons of water would be raised per minute 
from a mine 600 ft. deep by an engine of 175 H. P. ? 

Ans. 1152 gals. 

9. A belt passing round two pulleys moves with a velocity 
of 10 ft. per sec. : find the H. P. transmitted if the differ- 
ence of tension of the belt above and below the pulleys is 

1100 pounds. Ans. 20 II. i\ 



160 WOEK AND ENERGY. 

10. Show that the dimensions of power or rate of doing 
work, that is, of Fs/t, are MU/T\ 

11. A shaft 14 ft. in diameter is to be sunk in gravel in 
10 days of 10 hours each. Taking the weight of the gravel 
at 100 lbs. per cubic ft., find the H. P. required. 

138. For determining the power developed by a steam- 
engine or other machine, the Prony Brake is used. The 
idea is to balance the -work done by the machine by a fric- 
tional resistance, compute this resistance, and thence find 
the power of the machine. The brake absorbs the work to 
be measured. 

Let O be a shaft of radius r, to which the brake AB is 
fastened. By means of the screws a, b, the friction of the 

Fig. 147 




brake on the shaft may be regulated. Suppose it adjusted 
so that the engine develops a friction f, just sufficient to 
balance a body of weight W placed at the end A of the 
beam. Then the moments of /and W about O must be 
equal, or 

fr = Wl. 

Suppose the shaft to revolve uniformly n times per min- 
ute. Then, assuming that the friction for uniform motion 
of the shaft is the same as at the point of just beginning to 
move, we have 

Work done in one min. = friction developed in n revs. 

= /X 3rfr X n 

= 27tnWl, 



POWER. 1G1 

If W is expressed in pounds and I in feet, then the 

H. P. = 27r«TT7/33000 

= 0.00019/* WX 

Similarly, in the mechanism shown in Fig. 131, Art. 122, 
if the H. P. transmitted by the belt and the velocity v of 
the belt per second were given, then the tensions P, Q of 
the belt would be found from 

H. P.= (P- Q)v/6dO; P = Q&* 

or, if the H. P. and the number of revolutions per minute n 
of the driving pulley were given, the tensions would be 
found from 

H. P. = (P— Q)7tdn/3S000, P = Qe* Q 

when d is the diameter of the pulley. 

A common case is when jn = 0.3, and from 0.4 to 0.5 of 
the smaller pulley is embraced by the belt. Then 6 varies 
from 0.S7T to n, and P = 2Q, nearly. Also, the 

H. P. = O.OOOObPdn, 
P being the tension of the belt on the taut side. 

139. In a machine, owing to friction between the pieces, 
part of the work done by the driving force is wasted, so that 
the resulting useful work clone is less than the total work 
done by the effort in the first place. We have, in fact, 

Total work = useful work + useless work. 

The ratio of the useful work to the total work is known as 
the Efficiency of the machine. Or, since the total work is 
given by the indicated horse-power (Art. 136) and the useful 
work by the braked horse-power (Art. 138) we may define 
efficiency to be the ratio of the B. H. P. to the I. H. P., or 

Efficiency = B. H. P. /I. H. P. 



162 WOftK AND ENERGY, 

Ex. 1. In testing a Corliss engine running at 100 revolu- 
tions per minute, the lever arm was 10^ ft., and the weight 
at A 2000 lbs. : find the H. P. developed. Ans. 400 H. P. 

2. "A C. and C. electric motor shows on a Prony brake a 
pull of 5 ounces on a one-ft. lever, that is, 2 ft. -pounds per 
revolution, or about T V H. P. at 1500 revolutions per min- 
ute." Cheek the conclusions in this statement. 

3. In a Corliss engine running at 100 pounds pressure 
and 100 revolutions per minute, the diameter of the cylin- 
der is 18 in., and length of stroke 42 in. If the brake was 
used on a pulley 6 ft. in diameter, and keyed to the engine 
shaft, find the friction on the face of the pulley. 

Ans. f = 9450 pounds. 

4. A G-ton fly-wheel on a 14-in. axle makes 90 revolutions 
per minute. Find the II. P. absorbed in friction, the co- 
effieient of friction being 0.1. Ans. 12 H. P. 

5. A steam hoist of 3 H. P. is found to raise a weight of 
10 tons to a height of 50 ft. in 20 min. How many ft.- 
pounds of work are wasted by friction in a day of 10 hours ? 

Ans. 29,400,000 foot-pounds. 

6. The tractive force of a consolidation engine is 10 tons: 
find the II. P. exerted in hauling a train one mile in 2 min. 

Ans. 1G00 II. P. 

7. A pumping engine of piston area 100 sq. in., steam- 
pressure GO pounds per sq. in., length of stroke 3 ft., and 
number of revolutions per min. 25, raises 500 gallons of 
water per minute a height of 50 ft: find the efficiency. 

Ans. 0.25, nearly. 

8. A traction engine weighing 5 tons hauls a load of 10 
tons at 8 miles an hour, the resistance being 20 pounds per 
ton : find the II. P. exerted. Ans. G.4 H. P. 

9. A train weighing 100 tons runs at 42 miles an hour on 
a level track, the resistance being 8 pounds per ton : find 
its speed up a 1$ grade (1 ft. rise in 100 ft.) if the engine 
power is unchanged. 

[Total resist. = 8 + 2000/100 = 28 pounds per ton. 
.*. 8 X 42 = 28 X x, and x = 12 miles an hour.] 

10. A traction engine weighing 5 tons can haul 15 tons on 
a level, the coefficient of friction being 0.02 : find the net 
load it can haul up a 1$ grade. Ans. 8^ tons. 



POWER. 163 

11. A train of 100 tons is hauled by an engine of 150 
H. P. The resistance is 14 pounds per ton: find the great- 
est velocity that the engine can attain. 

Ans. 60 miles an hour, nearly. 

12. Check this statement : "55 pounds mean effective 
pressure at 600 ft. piston speed gives 1 H. P. for each sq. 
in. of piston area." 

13. Prove H. P. of an engine = SNA P / 33000, where 
S= stroke in ft., N = number of strokes per mm., A = 
area of piston in sq.in.,P = mean steam pressure in pounds 
per sq. in. of piston area. 

14. Find the work done per hour at the crank-pin of an 
engine revolving 40 times a minute and acting against a re- 
sistance of 7000 lbs., the radius of the crank being 18 inches. 
(See Fig. 119.) 

Ans. 2 7t x H X 7000 X 40 X CO ft.-pounds. 
14a. In the Strong locomotive 444, built for the L. V. 
E. E., the cylinders are 20 in. in diameter, the stroke is 24 
in., and the diameter of the driving wheels 62 in. At 160 
lbs. steam -pressure per sq. in. find the work done at each 
stroke. 

Ans. - X 20 2 X 2 X 160 foot-pounds. 

14:b. Find the tractive force P of the engine. 
[In each stroke of the piston the drivers make a half 
revolution. Hence, there being two cylinders, 

^ X 20 2 X 24 X 160 = | X 62 X P, and P is found.] 

15. Show that the cylinder diameter of an engine that 
will produce n horse-power at a piston velocity of s ft. per 
minute under a mean effective pressure of p pounds per 



SI 



sq. in. is 210 y — inches, nearly. 

16. The driving pulley (Fig. 131) runs at 100 revolutions 
per minute, and is 2 ft. in diameter. The engine is 3 II. P. 
Find the tensions of the belt ii' { of the circumference of 
the driving pulley is covered, and the coefficient of friction 
is 0.3. 



164 WORK AND ENERGY. 

17. In the California Street cable road, San Francisco, 
the total H. P. required to haul the cable alone is 84, and 
the speed is six miles an hour : find the total pull transmitted 
by the driving drum. 

140. Energy. — We have seen that when the forces acting 
on a body are in equilibrium, or the body moves with a 
uniform motion, the sum of the works done is zero, that is, 
the work done by the resultant force or effort is equal to 
that done on the resistance. Now the action of a force is to 
cause acceleration. If, theu, the motion is uniform, the 
acceleration caused by the effort is balanced by the equal 
and opposite acceleration caused by the resistance. But if 
the acceleration caused by the effort exceeds that caused 
by the resistance, velocity is gained, and the motion is not 
uniform. We proceed to inquire as to the work necessary 
to be done in order to change the velocity of a body of mass 
m from suy u ft. per sec. to v ft. per sec. Let F denote the 
effort or acting force, a the acceleration produced, and sthe 
distance passed over in the line of action of the force; then 
(Art. 13) 

as = W — %u\ 

But (Art. 34) F — ma. 

Hence, eliminating a, Fs = \mi^ — \mu*. 

Now Fs is the work done by Fin passing over a distance 
8 in its line of action, and therefore a mass m in having its 
velocity changed from u to v feet per second must have 
Imr' 2 — \mi? units of work done on it. 

If the force F does not act uniformly, we have, from Art. 
34, 

F= m(Ts/dt\ 
and 



. • . / Fds = I m -7T»ds = \mv 



ENERGY. 165 

or, the work done depends on the initial and final velocities, 
and is independent of the intermediate velocities. 
If u — 0, or the body starts from rest, 

Fs = \mv % , 

and therefore the work clone in giving a body of mass m a 
velocity v is \mc % units of work. The force F which will 
generate a velocity v in acting through a distance s will 
destroy the same velocity if acting through the same dis- 
tance in the opposite direction ; in other words, the body by 
virtue of its velocity v can do a work Fs units in giving up 
that velocity and coming to rest. This capacity which the 
body possesses of doing work in consequence of its velocity 
known as its vis-viva or Kinetic Energy. Hence the meas- 
ure of the kinetic energy possessed by a body of mass m and 
velocity v is ^mv 2 units of work. In acquiring the velocity 
v by the work done on the body energy may be said to be 
stored in it, to be restored Aw doing work as it parts with 
this velocity and returns to its original condition. We may 
therefore state the general relation 

Fs = i'inv' i — \mi? 

in the form: If a body or system of bodies with configura- 
tion remaining the same is in motion under the action of 
force, the work done in passing from one position to another 
is equal to the corresponding change of the kinetic energy. 
This is called the Principle of Kinetic Energy. 

Ex. 1. Find the work done in stopping a 100-lb. shot 
moving with a velocity of 1000 ft. per sec. 

Ans. 1,562,500 ft. -pounds. 

2. Find the force exerted in stopping a train of 250 tons 
in 1000 ft. from a velocity of 30 miles an hour. 

Ans. 15,1 '25 pounds. 

3. A shot pierces a target of a certain thickness // : show 
that to pierce one of 4 times the thickness twice the veloc- 
ity is necessary, tymv^h — $mv t * x4//. . • . i\ = 2v r \ 



166 WORK AND ENERGY. 

4. A blacksmith's helper using a 16-lb. sledge strikes 20 
times a minute, and with a velocity of 30 ft. per sec: find 
his rate of work. Ans. 3/22 H. P. 

5. A stone is thrown with a horizontal velocity of 50 ft. 
per sec. : find the velocity with which it strikes the ground 
which is horizontal and 6 ft. below the point of projection. 

Ans. 53.7 ft. per sec. 

6. Show that to give a train a velocity of 20 miles an 
hour requires the same energy as to lift it vertically through 
a height of 13.3 feet. 

7. A hoisting engine lifts an elevator weighing 1 ton 
through 50 ft. when it attains a velocity of 4 ft. per sec. 
If the steam is shut off, how much higher will it rise ? 

Ans. 200 °* 4> = 2000;/ X dist. 

8. In (7) find the time of rising 50 ft., supposing the mo- 
tion uniformly accelerated, and also find the H. P. of the 
engine. Ans. 25 sec; 7.3 H. P. 

9. Show that the energy stored in a train of weight W 
lbs. and moving with a velocity of V miles per hour is 

WV*/30 foot-pounds. 

10. A train of 100 tons is running at 30 miles an hour up 
a 2$ grade: find the H. P. required, the resistance on a level 
being 10 lbs. per ton, due to axle friction chiefly. 

Ans. 400 H. P. 

lift. In the Strong locomotive (L. V. E. R 444) running 
on the level at 30 miles an hour the tractive force is 8 tons. 
Taking the resistance of friction as 10 lbs. per ton, find the 
number of 20-ton cars that can be hauled if engine and 
tender weigh 100 tons. Ans. 75 cars. 

lib. Find the number that would be hauled up a 2% 
grade. Ans. 11 cars. 

lie. Find the H. P. exerted in the former case. 

Ans. 1280 II. P. 

12a. An engine exerts on a car weighing .20,000 lbs. a 
net pull of 2 lbs. jjer ton : find the energy stored in the 
car after going 2% miles. Ans. 264,000 ft. -lbs. 

12b. If shunted to a level side-track when the frictional 



ENERGY. 167 

resistance is 10 lbs. per ton, find how far it will run before 
coming to rest. Ans. 264,000/10 X 10 ft. = \ mile. 

12c. If shunted on a side track with a 1$ grade, how 
far will it run before coming to rest ? 

Ans. 264,000/(20 + 10) 10 ft. = \ mile. 

Vld. If there are brakes on half the wheels, and these are 
applied with a pressure of half a wheel load, how far will 
the car run up a lfo grade, the coefficient of friction be- 
tween wheel and brake-shoe being 0.2. 

[Total resist. = brake -|- grade + fric. = 130 lbs. per ton. 

Ans. 203 ft., nearly.] 

13. In the Westinghouse brake tests (Jan. 1887) at 
"Weehawken a passenger train moving 22 miles an hour on 
a down grade of 1$ was stopped in 91 ft. There was 94$ 
of the train braked. Taking the frictional resistance as 8 
lbs. per ton, find the net brake resistance per ton and the 
grade to which this is equivalent. 

[The brake has to overcome the energy due to the ve- 
locity and the lesistance due to the grade, but is aided by 
the resistance due to friction. Hence brake resist, per ton 
= 2000 X 22 2 /(30 X 91)' +.20 — 8.= 367 pounds. 

As only 94$ was braked, we have the net brake resist, 
per ton = 367/0.94 = 390 pounds, which is equivalent to a 
grade of 390/20 (= 19.5) per cent.] 

14. The tractive force of an engine is P tons. If the 
weight of engine and train is W tons and the frictional 
resistance n lbs. per ton, show that in going up an a% grade 
the velocity acquired in t seconds from rest will be Qgt ft. 
per sec. and the energy 0.5 WQfgt* ft. -tons, where 

Q = P/ W - a/100 - m/2000. 

141. The term energy arises from the consideration of 
muscular exertion in the first place. In doing what is 
called work in ordinary language we recognize that effort 
is needed, and that exhaustion follows after a time. It is 
necessary to store up work-capacity or energy by consum- 
ing food in order to be able to continue doing work. 

The same idea is extended to machines where the force 
exerted by the expansion of steam, by water in motion, by 
air in motion, etc., produces effects which can also be ob- 



168 WORK AND ENERGY. 

tained by muscular exertion directly applied. The machine 
is then said to be doing work, and from analogy the term 
agent is applied to it as well as to man. The agent thus 
forms, as it were, the converter of energy, whether of food, 
or of coal, etc., into work.* The energy existing in a stored- 
up condition is ready to be called on, and is hence known 
as Potential Energy. 

Suppose we call on the potential energy of steam to give 
motion to operate a mechanism, as for example a locomotive. 
The energy of the steam has enabled the locomotive to 
overcome the resistance offered by the friction of the 
wheels, the resistance of the air, etc., and to attain besides 
a certain velocity. If the steam be shut off the locomotive 
will not at once come to rest, but will continue for a time 
to overcome resistance as before. The energy communi- 
cated in excess of that required to overcome the resistances 
is not lost, but is stored in the form of motion. The energy 
existing in virtue of the motion, which thus continues to do 
work until exhausted, is, as stated above, called Kinetic 
Energy. 

It would thus seem that the energy of motion is produced 
at the expense of energy stored, and conversely the energy 
of motion may be made to do work of some kind to be 
stored in some form or other. The subject is a very large 
one, and we cannot go into it in detail. It is sufficient to 

* "Now. Buckland," said Stephenson, "I have a poser for you. 
Can you tell me what is the power that is driving that train ?" 
"Well," said the other, " I suppose it is one of your big engines." 
" But what drives the engine ?" " Oh, very likely a canny Newcas- 
tle driver." "What do you say to the light of the sun?" "How 
can that be?'' asked the doctor. " It is nothing else," said the en- 
gineer; '* it is light bottled up in the earth for tens of thousands of 
years — light absorbed by plants and vegetables being necessary for 
the condensation of carbon during the process of their growth, if it 
be not carbon in another form; and now, after being buried in the 
earth for long ages in fields of coal, that latent light isag.-iin brought 
forth and liberated, made to work as in that locomotive, for great 
human purposes." — Smiles. 



POTENTIAL. 169 

state, that, as a result of observation and experiment, the 
conclusion is arrived at, that energy may be transferred from 
one form to another, but can neither be created nor de- 
stroyed. In a word, energy is indestructible; or, as it may 
be expressed: The total energy of any body or system of 
bodies is a quantity which can neither be increased nor 
diminished by any mutual action of these bodies, though it 
may be transformed into any of the forms of which energy 
is susceptible. 

This principle, known as the Conservation of Energy, is 
the general principle premised at the beginning of this 
chapter. It involves the laws of motion and the principle 
of work as special cases, and consequently on it may be 
made to rest the whole subject of mechanics. 

142. Potential.* — An important application of the prin- 



* A graphical illustration of the fact that in the case of two elec- 
trified spheres the potential function is a measure of work, is given 
by Prof. A. M. Mayer as follows: 
Suppose an electrically charged 
sphere fixed in space with its centre 
at 0, and that another sphere charg- 
ed with a unit of similar electricity 
is pushed towards from an infinite 
distance along the line OX, and 
that the electric strain on the mov- 
ing sphere causes (without work) a 
vertical rod to slide out of its top 
in proportion as the stress between 
the spheres increases. As the sphere 
progresses along OX it will thus 
mark at each point of its progress 
the repulsive force existing between it and the fixed sphere. The end 
of the sliding-rod during the motion of the sphere from X towards 
will have traced out the curve DFCO, whose ordinates are as the 
inverse squares of their distances from 0. 

The potential at any point reached in the progress of the charged 
body towards = work done = resistance overcome in pushing body 
from infinite distance to that point; and this work done is measured 
by the sum of the resistances at each point of the path X length of 
path. But this product is equal to the area included between the 




170 



WORK AND ENERGY. 



ciple of work is the determination of the work done dur- 
ing the passage of a body of given mass from one position 
to another under the action of a force of attraction or of 
repulsion on the body. The region in which the force acts 
is called the Field of Force. We shall confine ourselves 
to fields of force, in which the law of force is that of the 
inverse square of the distance (Art. 110), and the forces 
themselves are gravitational, electrical, or magnetic. 

Thus suppose a particle of mass m placed at to exert 
an attractive force on a unit mass 
in its motion from A to B, the 
law of force being that of the 
inverse square of the distance. 
The forced of attraction between 
the particles is cm/r 2 , when they 
are at a distance r from one an- 
other, c being a constant. The 
work done by the force as the 
moves the indefinitely small distance CD in its 

in 
c T)Tp X ^^ 3 wnen CE is perpendicular to OD. 

Or putting OD = x, DE = dx, and noting that the force 




ordinate (say B) of path, the axis of X, and the curve, both indefin- 
itely extended; or say CBAD. 
The equation of the curve is 



y — a/x 2 . 
Area ABGD, indefinitely extended, = 



/■•*=/■ 



adx/x* = a/x. 



Or area indefinitely extended, which represents the work, is in- 
versely as the distance of ;?/, the bounding ordinate of area, from 
or V = Q/d, when Q = quantity and d = distance of the centres 
of the two charged spheres. 



POTENTIAL. 171 

being attractive dx is negative, this may be written 

— c — 2 dx. Hence the work done in bringing the particle 
from A to B is 



r>dx (i i\ 

— cm — cm , 

da x \r a) 



. when OA = a, OB = r. 

If the starting-point A is at an indefinitely great distance 
from the source of attraction 0, then 1/a = zero, and the 
work done is c?n/r, a result independent of the form of the 
path AB. 

The work necessary to bring a particle of unit mass from 
a position of zero attraction, that is, from an indefinitely 
great distance, to a position B, where the force of attrac- 
tion is finite and the distance from the attracting particle 
r, is called the Potential at the point B. It is usually 
denoted by the letter V, so that 

V= cm/r, 
when r is the distance of B from 0. 

If the field at consists of several particles of masses 
m 1 , m 2 , . . . , with distances i\ , r 2 , . . . from the unit mass 
B, the potential of B will be 

chA. + ? »A + - • •)• 

It follows that if V x denote the potential at any other 
point C the work done in bringing the particle from C to 
B by any path is V — V 1 . 

143. Conceive a surface at every point of which the 
potential has the same value. Such a surface is called an 
Equipotential Surface. If a particle is moved from any one 
point to any other point on this surface no work is done. 
Thus, if a railroad track is located on a " level " no work is 
jdone against the force of gravity in hauling a train on this 
track. 



172 WORK AND ENERGY. 

An equipotential surface for the force of gravity could be 
determined by finding the places at which the same mass 
would compress a spring the same amount. Roughly, it 
would be a sphere concentric with the earth. The work 
done on any body by the force of gravity in falling to the 
earth from. any point of this equi potential surface would be 
the same in amount. So also the work done in falling from 
one such surface to another would be constant in value. 
This suggests a method of finding the relation between the 
resultant force R and the potential V at any point. For 
suppose the potential at points A, B situated on two equi- 
potential surfaces to be V A , V & respectively. The result- 
ant force R at A (as force of gravity on a falling body) 
will be in a line AC, normal to the surface at A. The 
work done in moving a particle of unit mass from A to B 
will be the same as from A to C, when 6' is the point in 
which the normal at A meets the second surface, and is 
equal to R XAC. But this work is by definition equal to 
the difference of potential at A and B, or V —V 
Hence 

RXAC= V-V. 

A B 

If the distance between the surfaces is indefinitely small 
( = dr), this may be written 

R = clV/dr, 

or the resultant force is the rate of change of potential at 
the point in question, the direction of the force beino- 
normal to the equipotential surface through the point. 

Potential being expressed in terms of work done, the 
unit of potential is the same as the unit of work, the foot- 
pound al or erg. 

The theory of the potential is of great use in magnetic 
and electrical investigations. 

Ex. 1. Show that the dimensions of potential are ML/T. 



POTENTIAL. 173 

2. Show that (theoretically) an equipotential surface 
could be determined by finding the points at which a pend- 
ulum beating seconds had the same length. 

3. To find the potential of a particle of unit mass on 
the line through the centre C and perpendicular to the 
plane of a uniform circular disc of radius a, thickness h, 
and density 6, the distance OC being = b. 

[As in Art. 110, 

mass of ring = %nr X dr X li X o\ 
Hence V= f*c X %7th6rdr/OD 

= f a c X 27tJiSrdr/ Vtf-$r* 

= 27tchd( Vb* -f a" - b).] 

4. In (3) find the potential at the centre of the plate. 

Ans. %7tchda. 

5. Find the potential at the centre of a circular wire of 
density 6 and indefinitely small thickness li. Ans. %7tcli8. 

6. Find the potential at any point within a spherical 
shell of mass wand radius a. Ans. cm /a. 

7. Find the potential at a point without a spherical shell 
of mass m, radius a, and distant b from the centre. 

Ans. cm/b. 

8. Assuming the earth to be a sphere of 8000 miles 
diameter, prove that the potential of the unit mass 1 lb., 
situated on the earth's surface is — 21,120,000 foot-pounds. 

9<7. In a series of concentric spherical equipotential sur- 
faces the distance between any two is proportional to the 
square of the geometric mean of the distances from the 
centre. 

9b. Hence show that at great distances from the earth's 
centre the pound mass must be moved over a long path in 
order to do a ft. -pound of work on it. 

9c. For example, at the moon, 210,000 miles distant from 
the earth's centre, find the shortest path. Ans. 3G00 feet. 



CHAPTER VII. 
KINETICS OF A RIGID BODY. 

144. The term rigid body is used in the sense already 
denned of a body regarded as composed of particles so con- 
nected that no part of the body can be moved relatively to 
any other part. 

As in the case of a single particle, we shall consider the 
nature of the motions of the particles of the body, without 
reference to the forces causing the motions. This forms a 
problem in kinematics. 

A rigid body is fixed in position by fixing three points A, 
B, (', not in the same straight line. These points determine 
a plane ABO with reference to which the positions of all 
points in the body may be defined. Hence the displace- 
ment of any point in the body may be determined by not- 
ing the change in the positions of these three points, since 
the point in question must keep in a fixed position relative 
to the three points A, B, C during the motion. 

If the two points A, B remain unchanged in position 
during the motion, the third point C must describe a 
circle about an axis through A, B, and the motion is one 
of rotation. If only one point A is fixed, the points B, C 
may be brought from their initial to their final positions 
B 1 , C 1 by two rotations. For by one rotation AB may be 
brought into the position AB X , and by revolving about AB 1 
as an axis the plane A B X C may be brought into the position 
AB 1 C 1 . If no point is fixed, the new position A 1 B X C 1 may 
be reached by a translation of A to A x , and by two rota- 
tions bringing A % BG successively into the positions A 1 B 1 

174 




PL AXE MOTION. 175 

and A 1 B l C 1 . Hence every motion of a rigid body is either 
a motion of translation or of rotation, or some combination 
of the two. 

The most important case is that in which the particles of 
the body move in parallel planes. f 

Such a motion is called plane mo- / B ' 

Hon. F »"g- 'so / 

Let us consider the motion of a / B 

line joining the points A, B in the / /\ & 

plane of the paper. A translation A / / / . ~X 

A-A 1 and a rotation through an ! ^--^ / // „^'" Bl 

angle B'A 1 B 1 will displace AB into 

the position A l B l . So far as the 

displacement itself is concerned, it 

makes no difference whether the 

translation and rotation occurred simultaneously, or not. 

But this displacement might have been produced by a 
rotation only. For bisect AA 1} BB 1 in a, b, and let the 
perpendiculars aO, bO intersect in 0. Then evidently 
OA = OA 1 , OB = OB 1 , and angle A OB =A 1 OB 1 . Hence 
the displacement may be produced by rotation about the 
point 0. 

It follows therefore that any plane motion may be regarded 
as a motion of rotation about a centre in the plane of the 
motion, or, in other words, that a translation and a rotation 
about an axis perpendicular to the direction of rotation may 
be combined into a single rotation giving an equivalent 
motion. 

145. In many cases the final displacement h difficult to 
arrive at. This is particularly the case in mechanisms 
where the connections of the parts are often very complicated. 
Besides, the final displacement may not define clearly the 
intermediate displacements. It is therefore necessary to 
- study the displacement from instant to instant of the mo- 
tion. 




176 KINETICS OF A RIGID BODY. 

Suppose two points A, B of a body to have any motion 
in the plane of the paper. The 
points A, B will each trace out 
a path. Consider A. The line 
joining two consecutive positions 
of A will give the direction of 
motion in the path. This line is 
the direction of the tangent to 
the path at A. Since an indefin- 
itely great number of curves may have a common tangent 
at a point, it follows that this tangent is quite independent 
of the form of the path. Hence for the instant we may 
consider the path to be a circular arc. The perpendicular 
A to the tangent will pass through the centre of the circle, 
and conversely, the direction of motion at A for the instant 
will be perpendicular to the radius of the circle. Hence 
the instantaneous motion of A is the same as if it took 
place in a circle with centre somewhere on AG. Similarly, 
the motion of B is the same as if in a circle with centre 
somewhere on BO. But is common to AG and BG. 
Hence the instantaneous motion of A and B, and therefore 
of the line AB, is a motion of rotation about a point G, 
which is called the Instantaneous Centre. An axis through 
perpendicular to the plane of the paper is the Instant- 
aneous Axis. 

The points A, B are any two points in the body. Hence, 
whatever the form of the body, and whatever its plane mo- 
tion, it is always possible to find a point such that for 
the instant the motion about it shall represent the actual 
motion, in other words, at any instant one point G is at 
rest, and the other points are moving in directions perpen- 
dicular to the lines joining them to this point. 

If in Fig. 151 the radii AG, BO do not intersect, the 
tangents to the paths at A, B are parallel, and the motion 
is a, motion of translation. The radii being parallel maybe 
said to intersect at infinity, and hence a motion of transla- 



INSTANTANEOUS AXIS. 



177 



tion may be regarded as a rotation about a centre at an in- 
finite distance. 

146. In general, the instantaneous centre will vary in 
position from instant to instant. The locus or path de- 
scribed by it is called a Centrode.* But in case the radii 
AO, BO continue to intersect in the same point 0, as the 
motion progresses the instantaneous centre becomes a 
permanent or fixed centre. For example, a wagon wheel 
revolves about the axle as a permanent centre, but with 
reference to the ground it revolves about the point of con- 
tact as an instantaneous centre. The path traced by the 
wheel on the ground is the centrode. 

Ex. 1. A ladder BO slides between a vertical wall and 
the ground, which is horizontal: find the 
instantaneous centre and the centrode. 

[The paths are along AB, AC. Hence 
the instantaneous centre is at the intersec- 
tion of the perpendiculars BO, CO. It 
is evident that AO = BO, the length of 
the ladder, and .'. is at a constant dis- 
tance from A. Hence the centrode is a 
circle, with A as centre.] 

2. A lever moves about a fulcrum : find the nature of the 
centre of motion. 

147. In the case of moving bodies rigidly connected to- 
gether, the determina- 
tion of the velocity of 
one with respect to an- 
other may be based on 
the preceding. For 
illustration take the 
ordinary steam-engine. 
The mechanism itself 
has been already shown 
in Fig. 119. Suppose 
we wish to find the 

* Term introduced by Cliilord (1845-1879). 




Fig. 153 




178 KINETICS OF A RIGID BODY. 

velocity of the piston P, relative to the crank-pin B. The 
velocity of the piston is the same as that of the extremity 
A of the connecting-rod AB. The velocity of the crank- 
pin B is the same as that of the extremity B of the con- 
necting-rod. Hence the relation sought is the same as that 
between the velocities of the extremities A and B of the 
connecting-rod. 

The bed-plate A C is fixed. The extremity A moves in a 
straight line PC, and the direction of motion being along 
PC, the instantaneous centre is in a line A at right angles 
to PC. The extremity B moves in a circle of centre C, and 
therefore the instantaneous centre is in the line CB. Hence 
the connecting-rod is for the instant in the condition of a 
wheel turning about an axis through 0, the intersection of 
OA and CB. Consequently, 

velocity A : velocity B — OA : OB, 

or the velocities are as the distances from the instantaneous 
centre. 

If therefore the velocity of one of the two, piston or 
crank- pin, is given, that of the other follows at once. Thus, 
suppo.se the crank-pin to have a velocity 10 ft. per sec. Lay 
off to scale a distance BH — 10 ft., and draw EG parallel 
to BA. Then, since 

HB : GA = OB : OA, 

we scale off GA as the velocity of the piston. 

By drawing the crank in different positions, and finding 
the corresponding positions of G, a curve will result, the 
ordinates of which will give the velocity of the piston through 
its stroke. 

Ex. 1. In the above example draw the complete curve of 
piston velocity on a scale of velocity 5 ft. per sec. — 1 in, 
and of dimensions 1 ft, = 1 in, 



ANGULAR VELOCITY. 179 

2. Find the height above the track of a point on a 6-ft. 
locomotive wheel running on a straight track, that has halt 
the velocity of the highest point of the wheel. 

Ans. 1.5 ft. 

148. Angular Velocity.— If the motion of the body is a 
motion of rotation about a fixed axis, each particle describes 
a circumference, whose centre is in the axis. Since each 
circumference is described in the same time, the velocities 
of the particles must be proportional to the distances of the 
particles from the axis, the greater the distance the greater 
the velocity. It is therefore clear that we can attach no 
meaning to the phrase " velocity of a body" as in the case 
of the motion of translation. In a word, we have a new 
kind of motion, and we must introduce new modes of 
measurement. 

Thus suppose the body to revolve about an axis through 
0, and that it moves through an angle 
AOa in t sec. Then the angle de- 
scribed in one second, or the Angular 
Velocity go, would be measured by 
AOa/t. This angle is described by 
every point in the body, so that one 
characteristic of rotation is that the 
angular velocity of every point is the 
same. 

The unit of angular velocity is naturally taken to be unit 
angle described in one second. The unit angle employed 
is the radian or the angle A Ob, whose arc Ab is equal in 
length to the radius A. 0. 

The angular velocity go will therefore be denoted by the 
number of radians described per second. Thus if a body 
revolves 60 times per minute, or once per second, the number 
of radians described per second is %n. 

The relation between the angular velocity go of the body 
and the linear velocity v of any particle A situated at a 




180 KINETICS OF A RIGID BODY. 

distance r from the axis of rotation follows at once. For 
the time of motion being /, we have 

oo = A Oa/t, arc Aa — vt. 

But 

A Oa : A Ob — arc Aa : arc Ah, 

or oot : 1 radian = vt : r, 

and oo — v/r radians, 

the relation sought. 

Ex. 1. Show that the dimensions of angular velocity are 
1/T. 

2. If oo is expressed in degrees, show that ? = 27rcyr/360°. 

3. A body makes n revolutions per second: show that 
the linear velocity of a particle 1 ft. from the centre is %7tn 
ft. per sec. and the angular velocity 27tn radians per sec. 

4. A belt passes over a pulley d ft. in diameter and mak- 
ing n revolutions per min. Find its velocity. 

Ans. ndn ft. per min. 

5. A wheel 4 ft. in diameter revolves 420 times per min. 
Find the angular velocity and the linear velocity of a point 
1.5 ft. from the centre. 

Ans. 147T radians per sec. : 21 n ft. per sec. 

6. The crank of an engine makes n revolutions per min. 
Its radius is r ft. Find the linear velocity of the crank- 
pin. Ans. 7trn/30 ft. per sec. 

7. In the driving wheel of a locomotive show that for an 
instant one point is moving twice as fast as the locomotive 
and in the same direction. 

8. Find the ratio of the angular velocities of the hour 
and second hands of a watch. Ans. 1/720. 

9. A locomotive is running at 45 miles an hour. The 
driving wheels are ft. in diameter, and the stroke is 2 ft. 
Find the piston velocity. A)is. 44/ it ft. per sec. 




EQUATIONS OF MOTION. 181 

149. Angular Acceleration. — Angular velocity, like linear 
velocity, may be uniform or variable. If variable, the rate of 
change is called the Angular Acceleration. The unit of 
angular velocity being one radian per sec- 
ond, the unit of angular acceleration is 
one radian-per-second per second. 

If the body rotates with a uniform an- 
gular acceleration a, the gain of circum- 
ferential velocity per second (or the linear 
acceleration) of a particle A at a distance r from the axis 
is measured by an arc Act equal to ar. Hence we may 
write 

linear accel. = ang. accel. x rad. 

The direction of the acceleration at A for an indefinitely 
small arc is normal to A 0, that is, along the tangent at A, 
so that we have 

tang, accel. =■ ang. accel. x rad. 

Hence, if m denotes the mass of the particle A, 

tang, force = mass X tang, accel. (Art. 34) 



Ex. When steam is shut off, the fly-wheel of an en- 
gine is making 90 rev. per min. If the coefficient of fric- 
tion is 0.1, find the time in which the wheel will come to 
rest. Ans. oOn sec. 

Thus far Ave have considered the kinematics of a rigid 
body. We now proceed to discuss its kinetics. 

150. Equations of Motion. — The result of the action of 
a force on a particle of a rigid body is different from 
that on the particle if free. For besides this force other 
forces act on the particle resulting from the action of 
the adjacent particles of the body on one another. The 



182 KINETICS OF A RIGID BODY. 

real acting force is therefore the resultant of the external 
and internal forces, and the motion is the same as if we 
considered the particle free, but subject to the action of a 
force equal to the resultant of all external and internal 
forces. To this resultant the name of Effective Force is 
given. 

Thus, suppose P to be a particle of a body of mass m 1 , 
F t the component in a certain direction of the external 
force impressed on it, and F x ' the component of the re- 
sultant internal forces in the same direction. If a 1 is the 
component of the acceleration in this direction due to the 
resultant of the forces F l , F x ', the effective force acting on 
the particle must be equivalent to m l a 1 . Hence 

F x + F' = m x a x . 
Similarly, F n -f- F 2 ' = m n a 2 . 



By addition, 

2F + 2F' = 2ma. 

But from the law of stress the internal forces occur as 
pairs of forces equal in magnitude and opposite in sense, or 
2F' = 0, and therefore 

2F — 2ma, 

or the system of impressed forces is in equilibrium with the 
system of effective forces reversed. 

Now if a is the acceleration of the centre of gravity of 
the body, then (Art. 90) 

a = 2ma/2m 9 

and . * . 2F = 2ma ; 

or the linear acceleration of the centre of gravity is the 
same as that of a particle of mass 2£nt acted on by a force 
2F, 



EQUATIONS OF MOTION. 183 

Consider now the rotation of a rigid body about a fixed 
axis under the action of external forces. 

Let the axis through be perpendicular to the plane of 
the paper, and let at an assigned instant F i , F 2 , . . . be the 
components of the external forces on the particles A, B, . . . 
acting at distances i\ , i\ , . . . from the axis, and parallel 
to the plane of the paper. Also, let F/, F 2 f , ... be the 
components in their respective planes of the internal forces 
on the particles A, B, . . . , and denote by R l} R 2 , . . . 
the resultant forces at A, B, . . . These resultant forces 
may be resolved into components along the tangent and 
radius to the paths of A, B, . . . The tangential accelera- 
tion is due to the tangential component. If, therefore, a 
denotes the angular acceleration of the body, the linear tan- 
gential acceleration at A is i\a, and the tangential force 
(or inertia-resistance) m x r x oc. Denote the normal compo- 
nent at A by N x . Similarly, we have m 2 r 2 a and N 2 at B, 
and so on for the other points. These are the effective 
forces. Taking moments about 0, we have, iip l9 p i3 . . .; 
V\ >P*> • • • I • • • denote the perpendiculars let fall from 
on F l3 F 2 , . . .; F x ', F* ; ...;... respectively, 

m x i\a X r, + mj\a X r 2 -j- . . . 

= f iPi + F iPi + . . . + F/p/ + f; v ; + . . . 

or 2mr*a = 2Fp + ^F'p'. 

But the internal forces consisting of pairs of equal and 
opposite forces, we must have ^F'p' — 0. Also, since a 
is the same for all the particles, 2mr 2 a = aJEmr ; and Ave 
have, finally, 

a2m?' 2 = ^Fp. 

The right-hand member of this equation is the ordinary 
expression for statical moment. The left-hand member is 
the sum of the moments of the inertia-resistances m, x r x a } 



184 KINETICS OF A KIGID BODY. 

m^r^a, . . . about the axis, and may therefore be appropri- 
ately called the Moment of Inertia* of the body about the 
axis. But the term moment of inertia is more usually ap- 
plied to the factor ^mr 2 , which is constant for the body 
considered and independent of the varying acceleration a. 
This factor is denoted by the letter 7, so that 

We may therefore write the equation 

ang. accel. = mom. of ext. forces/mom. inertia. 
The relations 

2F = a2mr, 

2Fj) = a2mr* , 

the first giving the acceleration of the centre of gravity in 
terms of the external forces and the mass, the second the 
angular acceleration about a fixed axis, are called the Equa- 
tions of Motion of a rigid body. 

Having found the accelerations from these equations, and 
being given the initial velocities, the velocities at the end of 
any given time and the distances passed over may be determ- 
ined. 

151. The conditions of equilibrium of a rigid body already 
found in Art. 88 follow at once. For for equilibrium the 
acceleration of the centre of gravity must be zero and the 
angular aeceleration must be zero. Hence, since a = 0, 
a = 0, we have 

2F=0, 2Fp = 0, 

or, the sum of the components of the external forces in any 
direction is zero, and the sum of the moments of the external 
forces about any ]Joint is zero. 

In fact all static problems may be regarded as limiting 
cases of kinetic problems, and may be treated accordingly. 

* The term was introduced by Euler (1707-1783). 



MOMENT OF INERTIA. 185 

152. Moment of Inertia, — It will be convenient in this 
place to give some examples of moments of inertia that are 
of frequent occurrence. We shall for simplicity and con- 
venience make use of the integral calculus. Indeed, the 
finding of moments of inertia is a problem of summation or 
integration, and is to be regarded as such, and not a mechan- 
ical question at all. This summation may, however, some- 
times be effected without the calculus, and as an illustration 
Ex. 1 is solved in both ways. 

First take the axis through the centre of gravity of the 
body* 

Ex. 1. To find the moment of inertia I of a thin uniform 
rod, mass 31, length I, about an axis OY pj g . 155 

through its centre 0, and at right an- 
gles to the rod. 

[Conceive the rod cut into elements 
of indefinitely small length dx, and let 

x be the distance of any one of these | x a .i 

elements from 0. Then moment of o H ~ ., 

inertia of element = ddx X x 2 when d is the linear density, 
and . • . Sdx the mass of length dx. Hence 



-A 



dx\lx = oT/12 = MP/12. 



Or thus : Suppose the rod divided into a large number 2n of 
equal parts l/2n. The distances of these parts from may 
be taken to be the distances of their centres of gravity from 
0, that is, 1/4:11, 31/4:11, . . . Hence, taking half the rod, 



e i{iJ +6 iiS + 




to n 


terms 


32^ r + 32 + 


. . to n 


tor 


ms) 




24 V 4n 7 











186 



KINETICS OP A RIGID BODY. 





Y 

< h 


Fig 


. 157 


h 




./■ 




X 
























= — -, when n is indefinitely great, 

and I = MP/12, as before.] 

2. A thin rectangular lamina or 
plate of breadth b and depth h, 
about an axis through its centre of 
gravity and parallel to h. 

[Conceive the lamina cut into 
strips parallel to the axis, and of 
breadth dx. Let x denote the dis- 
tance of one of these strips from 
the axis, S its density ; its mass is 
6h X dx. Hence 



« 7 2 



67i X dx X x 2 = ^Shb 3 = JiMb*.] 



3, If in (2) the axis is parallel 
to the side b, show that / = 
Mh*/l%. 

4. A thin circular plate, radius 
r, about a diameter YOY' as axis. 

[Conceive the plate cut into 
strips parallel to the axis, and of 
breadth dx. Then the equation 
to the circle being x 2 -\- if = r 2 , 
the length of a strip at a distance 
x from is 2 Vr 2 — x\ and mass 
of strip = S X 2 Vr 2 - x\lx. 
Hence 




26V Vr 2 - x 2 dx = &7cr*/± = Mr*/4. 



5. A square plate of side a about a diagonal. 

Ans. 1= Ma 2 /\2. 
G. A hexagon of side a about a diagonal. 

Ans. 1= bMa*/±, 



MOMENT OE INERTIA. 



187 



7. A triangle of base b, height //, about an axis through 
its centre of gravity and parallel to the base. 

Ans. I = ^ 3 /36. 

Fig. 159 
-| I 1 r—| Fig- '60 






8. A channel iron or I-iron, base b, depth h, thickness of 
web b x> depth of web h l3 about axis h through 0. of G-. 

Ans. l=\bh 3 - (b- b^y/12. 

9. A T-iron with dimensions as in Fi<r. 160 about axis 
through C. of G. Ans. I = {U* + bJi^/W. 

153. The form of the expression for the moment of inertia 
about a fixed axis or a fixed point, 2mr 2 , shows that we may 
define it as the sum of the products of the masses of the 
particles of a rigid body into the squares of their distances 
r from the fixed axis or from the fixed point.* If the body 
consist of an indefinitely thin plane lamina, and be referred 
to rectangular axes OX, OY in its plane, we may write 
r 2 = x 2 -j- y 2 and 

I=2m{z 2 ±y 2 ) 

for the moment relative to or to an axis through per- 
pendicular to the plane of the lamina. But ^mx 2 , ^mif 
are the moments of inertia relative to the axes OY, OX, 
respectively. Hence, for a body in the form of an indefin- 
itely thin lamina, if I x , 7 a denote the moments of inertia 
relative to two rectangular axes OX, OY, in the plane, 
and / the moment relative to an axis through perpen- 
dicular to the plane, we have 

/=/, + /, • 

* This. is the usual definition. It fails to show the significance of 
the term moment of inertia. (See Art. 150.) 



188 KINETICS OF A RIGID BODY. 

Ex. 1. A rectangular lamina, breadth b, depth h, about an 
axis through its centre of gravity and perpendicular to its 
plane. Ans. I = M(b* + A 2 )/12. 

2. A circular plate, radius r, about an axis through its 
centre and perpendicular to the plate. Ans. I = Mr 2 /2. 

3. A ring of radii r\, r a about an axis through its centre 
and perpendicular to its plane. Ans. I = M(r* -\- r 2 ) /2. 

4. A sphere, radius r, about a diameter as axi.s. 
[Conceive the sphere divided into plates of width dx 

by planes perpendicular to the axis. Let the distance of 
any plate from the centre be x. Then radius of plate 
= V r 2 — x\ and mass of plate = 671(1^— x*)dx. Hence, 
from Ex. 2, 

/ = f * \d7t{r* - x')dx X (r 2 - x% 

= f % d7rr' = iMr\] 

5. Show, by differentiating the result of Ex. 2, that the 
moment of inertia of an indefinitely thin ring of radius r 
about an axis through its centre and perpendicular to its 
plane is Mr 2 . 

0. Show by differentiating the result of Ex. 4 that the 
moment of inertia of an indefinitely thin spherical shell 
about a diameter is 2 J/r 2 /3. 

154. The moment of inertia of a body about an axis not 
passing through the centre of gravity may be readily re- 
ferred to a parallel axis through that 
point. For suppose the two parallel 
axes through a point and the centre 
of gravity G to lie in a plane perpen- 
dicular to the plane of the paper. 
Take G as origin, the plane of the 
paper the plane of X, Y, and OG X the 
axis of X. Let x, y denote the coor- 
dinates of any particle P of mass m. Call the .distance 





Fig. 161 




,/P 


Y 


// 


/ 


// 



MOMEXT OF IXEKTIA. 189 

GO = d. Then if /denote the moment of inertia about 
an axis through 0, we have 

= 2m(if-\- x*)+ 2d2mx + d?2m, 

since the distance d is constant. 

Of the three expressions in the right-hand member, the 
first is equal to I x , the moment of inertia about G; the sec- 
ond is equal to zero, since G is the centre of gravity (Art. 
90) ; and the third is equal to Md 2 , where M is the total 
mass. Hence 

I=I t + Md\ 

or the moment of inertia about any axis is equal to the mo- 
ment of inertia about a parallel axis through the centre of 
gravity, together with the product of the mass into the square 
of the distance between the two axes. 

Ex. 1. Given that the I of a rod of length I about an 
axis through its centre is Ml* /12 (Ex. 1, p. 185), show that 
about one end it is i/7 2 /3. 

2. A rectangular lamina about A GB Fia 162 
has I=Mh 2 /12 (Ex. 3, p. 186): show 
that about CD it is = Mh 2 /3. 

3. A circular lamina, radius r, about 
its centre has / = Mr 2 /2 (Ex. 2, p. 
188) : show that about any point in the 
circumference / = 3Mr*/2. 

4. A triangular lamina of base b, 
height //, about an axis (1) coincident 
with the base, (2) through the apex and parallel to'the base. 

Arts. b¥/\2 ; bh"/\. 

5. Prove that i" is the same for all parallel axes situated 
at equal distances from the centre of gravity. 

6. Of all parallel axes the / of that which passes through 
the centre of gravity is the least. 

155. Radius of Gyration. — The general expression for 
the moment of inertia of a body relative to an axis is 





C I 


) 


A 


G 


B 


h 


b 





190 KINETICS OF A ltlGID BODY. 

2mr*, the sum of the products of the mass of each particle 
into the square of its distance from the axis. Kow instead 
of the particles being distributed in this way, we may con- 
ceive them concentrated into a single particle of mass M 
equal to the whole mass, and at such a distance k from the 
axis that the product Mk* — 2mr*. To this distance k 
the name Radius of Gyration * is given. 

If the axis passes through the centre of gravity, we write 
Mk* = ^Emr* where k l is called the principal radius of 
gyration. 

The relation between k and k x follows at once from the 
preceding article. For 





Smr* = 2 an-; + Ma% 


or 


Mk" = Mk; + Ma\ 


or 


¥ = k; + a\ 



the relation sought. 

Also, since J 1 — Mk;, the value of k\ is at once found. 

Ex. 1. Show that the principal radius of gyration is a 
minimum radius for parallel axes. 

2. For a straight line of length I with reference to its 
centre, show that k* = T/12. 

3. For a rectangle of breadth /;, depth h, with reference 
to its centre of gravity, show that k* = (U 1 + A 2 )/12. 

4. For a circle of diameter d with reference to its centre 
show that k; — d' 2 /lQ. 

5. For a triangle of base o, height h, about an axis 
through its centre of gravity parallel to the base, show that 
k; = h*/18. 

6. Find fc' for a right cylinder about its axis, r being 
the radius of cross-section. 

[Since the cylinder maybe conceived to consist of an in- 
finite number of plates, each of which has the same radius 
of gyration with respect to an axis through the centre and 
perpendicular to their planes, the radius of gyration of the 
cylinder is the same as that of any plate, aud .-. k; = r*/2.] 

* Called also radius of inertia. 



THE PHYSICAL PENDULUM. 



191 



7. For a right cylinder about an edge k* — 3r 2 /2. 

8. For a hollow cylinder, inner radius r, , outer radius i\ , 
relative to its axis, k 2 = (r^ -f- r 2 )/2. 

9. For a vertical cylinder, radius r, length I, about a 
horizontal axis through its centre of gravity show that 
k 2 = r'/± + f/12. 

10. For a rectangular prism of dimensions a, h, c about 
the edge a, show that h* = (b 2 + c 2 )/12. 

156. We shall now give some applications of the general 
equations of motion to special cases in which the axes of 
rotation are fixed, and in which they are instantaneous. 

(a) The Physical Pendulum. — In Art. 73 was considered 
the problem of the time of oscillation of a heavy particle P 
suspended from a fixed point by a weightless rod, and acted 
on by the force of gravity. This problem is purely hypo- 
thetical, as no such apparatus can be constructed. 

But just as a rigid body is regarded as built up of parti- 
cles joined together, so an' actual or 
physical pendulum may be regarded 
as composed of simple pendulums 
whose oscillations so act on one 
another as to result in a common 
oscillation. The duration of this 
oscillation will give the length of 
a simple pendulum which fulfils 
exactly the same conditions of mo- 
tion. 

Suppose the physical pendulum to 
be a body of any form, and let the 
two be placed side by side, and both 
swung through an angle 6. The 
external forces on the physical pend- 
ulum are the weight Mg acting at the 
centre of gravity G, and the reaction 
R at the point of support 0. Hence, since the moment of 



Fig. 163 




192 KINETICS OF A EIGID BODY. 

R about C is nil, we have 

ang. accel. — Mg X GH/I 

= Mgh sin 6/1, 

where /is the moment of inertia with respect to the axis 
through C, and li is the distance CO. 

For the simple pendulum, Mg being the weight of the 
particle at P, 

ang. accel. = Mg X PQ /mom. inertia 

= Mgl sin 6/ Ml* 

— g sin 6/1. 

But the angular acceleration being the same in the two 
cases, 

Mgl t sin 6/1 — .g sin 6/1, 

or / = 1 /Mil, 

which gives the length of a simple pendulum with the same 
motion as the compound. 

Hence the time of oscillation / of the compound pendu- 
lum is (Art. 73) 

t = 7t VT/g — re Vl/Mgh. 

A point D at a distance I from C, the point of suspension, 
is called the Centre of Oscillation,* for the reason that the 
time of oscillation of the whole pendulum is the same as 
that of a simple pendulum of length /and swinging about C. 
Denote the distance DG by h, so that li -\- k — 1. 

Suppose now the pendulum inverted, and suspended at 
D instead of at C. The time of an oscillation is 

TtVTTM^k, 

* First determined by Huygens. 



THE PHYSICAL PESDULUM. 193 

where I' is the moment of inertia about D. To find the 
centre of oscillation if the time of oscillation is the same 
as if suspended at C, we must find the length \ of the sim- 
ple equivalent pendulum. We have 

/=/, + Mh\ 

r = i 1 -\-Mk\ 

where I 1 is the moment of inertia about G (Art. 154). 
But I=Mhl, I' = Mkl l3 

and . * . by an easy reduction 

or the centre of oscillation in the latter case is the centre of 
suspension in the former. 

Hence the points of suspension and of oscillation can be 
interchanged without changing the time of oscillation, and 
appropriately therefore a pendulum with points of suspen- 
sion situated as C, D is known as a Reversion Pendulum. 

The principle of reversion is employed to determine the 
length / of the simple equivalent pendulum experimentally. 
Theoretically, I can be found from 1 = Mid, but practi- 
cally it is difficult to make the measurements required 
by that equation. Hence the experimental method is gen- 
erally employed.* A pendulum of given form is suspended 
from a point C (on a knife edge) and caused to oscillate. 
By trial another point D is found, from which if it is sus- 
pended it will oscillate in the same time. The distance 
between C and D is carefully measured, thus giving the 
length I, 



194 KINETICS OF A RIGID BODY. 

Again, the value of / being known, and the position of 
the centre of gravity found as by balancing on a knife edge, 
we have at once the moment of inertia of the pendulum 
from 

/ = MM. 

This method of finding moments of inertia is particularly 
useful with solids of irregular figure, or if not perfectly 
homogeneous throughout. The solid is mounted as a pend- 
ulum. 

Still further : The length / being known and the time of os- 
cillation being observed, we have from the relation t = 7t Vl/g 
the value of g, the acceleration due to gravity. 

The length of the seconds pendulum was used in England 
for a time as the standard of length, but was afterward 
abandoned for a certain brass rod called the standard bar, 
for the reason that several of the elements of reduction of 
pendulum experiments are doubtful. 

Ex. 1. A rod of length I is suspended at one end, and 
caused to oscillate : find the length of the equivalent simple 
pendulum. Ans. 21/3. 

2. If the rod in Ex. 1 is suspended at J of its length from 
one end, find the time of an oscillation. 

3. If a pendulum is suspended at the principal centre of 
gyration, prove that the time of oscillation is a minimum. 
[For t — n V(h 2 + k*)/gh, and i is a min. when h = &,.] 

o!s Fig. 164 (0) The Steam Engine. 

j \ — To illustrate motion 

\ / about an instantaneous 

\ i/q axis, we shall confine our- 

\b^-— \ selves to the steam en- 

^^<P\ >v gine, and the still more 

__5 zfc^^™^4__._j^ j simple case of the rolling 

A S V J disc or sphere. 

^ — ^ The relation between 
the piston pressure P and the crank-pin resistance Q, when 



STEAM ENGINE. 195 

the connecting-rod is inclined at any angle, lias already been 
solved in i\.rt. 109, bnt may be solved more simply by aid 
of the instantaneous centre. 

For the velocity v\ of the cross-head is to the velocity v 2 
of the crank-pin as the instantaneous radii OA, OB, directly 
(Art. 147). But by the principle of work, P Xv x — Qxv^ 
Hence 

P x OA = Q x OB; 

or, P is to Q as the instantaneous radii inversely. 

The value of Q for a given piston pressure will thus vary 
according to the position of the connecting-rod. It may 
be represented graphically, as in the case of the indicator 
diagram (Art. 136). 

The average of the values of Q for a complete revolution 
of the crank corresponding to a given piston pressure P 
will be found by equating the work done by each of the two 
forces. We have, if r is the radius of the crank arm and 
S the length of the stroke, 

Q X 27tr = P X 28. 

But S = 2r; 

.-. 7tQ = 2P, 

the relation required. 

Ex. 1. In a Norris engine the diameter of the cylinder is 
14 in., and the steam-pressure 75 pounds per sq. in.: find 
the average value of the force acting on the crank-pin. 

Arts. 2 X 7 2 X 75 pounds. 

2. In (1) find the force acting when the crank stands at 
60°, and the ratio of the connecting-rod to the crank is 54. 

3. In a steam riveting machine the piston pressure P Is 
applied at the joint a, and the rivet squeezed between the 
jaws c, d. Find the relation between P and the force Q ex- 
erted on the rivet. [The instantaneous centre is at 0, 



196 



KIXETICS OF A RIGID BODY. 



where ba and the perpendic- 
ular through c to the sliding 
surface S intersect. Hence 

P xaO= Qx cO. 

As a approaches g, cO dimin- 
ishes; and when a reaches g, 
Q becomes indefinitely great. 
Hence the advantage of the 
apparatus in that an enormous 
pressure may be produced 
by a moderate force acting 
through a small distance. 

This is an example of the Tog- 
gle Joint, a mechanism of very 
considerable importance. It 
is applied, for example, in the 
Westinghouse air brake on locomotive drivers, in cider, 
oil, and other presses, etc., etc. In Fig. 166 is shown part 
of a power screw oil press. 

Fig. 166 



$p 


Fig. 165 








g'r 


\« 


/ - 


7\ 




> 




1 // 


/ \ 






s 


\/( 




\ 




l'\ 




LfXLJ 
" 1 







\ 

\ 
\ 
\ 
\ 

\ 




ENERGY OF ROTATION". 



197 



4. A cylinder rolls down an 
inclined plane of height h from 
rest : find its velocity v at the 
bottom. 

[ is the instan. centre ; r 
the radius of the cylinder. 
The forces acting are the wt. 
Mg and the reaction N. 



. \ ang. accel. a = Mgr sin Q/\Mr i — %g sin I 
. \ linear accel. of centre = ra = \g sin d, 




sfmg 



and v 2 = 2as ■ 



\g sin e x ^— d = igh.] 



5. A spherical shot rolls down a plane 70 ft. long and 
inclined at 30° to the horizon : find its velocity at the 
bottom. Ans. 40 ft. per sec. 

6. A sphere will roll and not slide down an inclined 
plane if the coefficient of friction is greater than -f tan a 
where a is the inclination of the plane. 

7. The kinetic energy acquired by a sphere in movie g 
from rest down a smooth plane is to that acquired by an 
equal sphere rolling down a rough plane of the same incli- 
nation and length as 7 to 5. 

157. Energy of Rotation. — The energy stored in a body 
rotating with a given angular velocity go will be found by add- 
ing together the energy of each particle. If v is the linear 
velocity of a particle and r its distance from the axis of 
rotation, we have v — gov, and the energy stored = ^mv 2 
= %niGo*r*. Hence the energy stored in the body is 
2$m(*>*r* where ^ is the symbol of summation. If n is 
the number of revolutions made per minute, gj = 27rn/60, 
and we have finally energy of rotation = ^l/ii&j-r 
= iloj* = In*/180, nearly. 

We shall confine ourselves to the cases of a circular disc 
and of a ring rotating about axes through their centres, per- 



198 KINETICS OF A RIGID BODY. 

pendicular to their planes. Common examples are a car- 
wheel and a fly-wheel. 

For a uniform disc of radius r, I = ^Mr 2 , and for a ring 
with r, , 7\ the inner and outer radii / = \M{r* + r 2 2 ), or 
if r„ r 8 are nearly equal, I = Mr m *, where r m is the mean 
radius of the ring. 

It is seldom that a wheel is in the form of a simple uni- 
form disc. In general the greater part of the weight is 
contained in a ring next to the rim of the wheel. This is 
more the case in a locomotive-wheel than in a car-wheel, 
and more in the fly-wheel of a stationary engine than in a 
locomotive-wheel. The two extremes are the uniform disc 
and the ring. 

In order to find readily the mass if of a fly-wheel, we 
notice that since 1 cub. in. cast-iron weighs 0.261 lb., we 
have, if A is the area of the cross-section of the rim in 
square inches, and d the diameter in inches, 

M = Ttd x A X 0.2G1 = 0.82dA lb., nearly ; 

and .-. I = 0.205 Ad*, nearly. 

158. In a stationary engine the fly-wheel is introduced to 
give steadiness. It does this by giving up when called on 
part of the energy stored in it.* To compute the dimen- 
sions of a wheel, we must decide on the greatest amount 
of energy that should be demanded of it, and also on the 
maximum range of velocity that can be allowed. Suppose 



* " The proprietor was showing to a friend the method of punch- 
ing holes in iron plates. He held in his hand a piece of iron f in. 
thick, which he placed under the punch. Observing after several 
holes had been made that the punch made its perforations more and 
more slowly, he called to the engine-man to know what made the 
engine work so sluggishly, when it was found that the fly-wheel 
and punching apparatus had been detached from the steam-engine 
just at the commencement of his experiment." — Babbaye. 



ENERGY OF ROTATION. 199 

we wish to call on it for 36000 ft. -pounds, and that the ve- 
locity may change from 48 to 52 revolutions per minute. 

Then /(52 2 - 48 2 )/180 = 36000# 

and J=162(%. 

If we decide to make the wheel 14 ft. in diameter, 

Mr* = 16200# 

and M = 5 -f- tons, 

whence the cross-section of the rim may be computed from 
the formula given above. 

159. In many cases a body possesses both an energy of 
translation and an energy of rotation, and their compara- 
tive amounts or their sum may be required. Take, for ex- 
ample, a railroad car in motion. Each wheel of the car acts 
as a fly-wheel in which energy is stored to be given out 
before the car comes to 'rest. If v is the velocity of the 
train, the tangential velocity gov of the wheel is equal to 
the velocity of translation v. Hence the energy stored in 
the wheel considered as a disc = %Igd 2 = |x J- Mr* X tf/f* 
= \Mv*, being one half that due to the forward motion 
of the wheels. Suppose the total (loaded) weight of the car 
to be 40000 lbs., and that the eight wheels weigh 4500 lbs., 
and the velocity is 30 miles an hour. Then the total energy 
stored in the car is 

£ X 40000 X 44 2 + i X 4500 X 44 2 foot-poundals, 
which may be reduced to foot-pounds by dividing by 32.2. 

Ex. 1. In a disc revolving about its axis the radius of gy- 
ration is nearly 0.7 of the actual radius. 

2. The rim of the fly-wheel of a Norris engine is 14 ft. 
in diameter, and weighs 11400 lbs.: find its I about the 
centre, 



200 KINETICS OF A RIGID BODY. 

3. The rim of a fly-wheel weighs 15 tons, and its diameter 
is 20 ft. ; the wheel makes 60 revolutions per minute : find 
the energy stored. Ans. 1,875,000 ft. -pounds. 

4. A fly-wheel of a tons wt. and b ft. diameter makes c 
revolutions per minute : find the energy accumulated. 

Ans. 0.087«&V ft. -pounds, nearly. 

5. The weight of a fly-wheel is W lbs., and its diameter 
d x inches. If it is making n revolutions per minute, find in 
how many revolutions it will be stopped by the friction of 
the axle if its diameter is d 9 inches and the coefficient of 
friction//. Ans. 7tn*d*/7200 Mff d* « 

6. Examine the following statement: " Every engineer 
knows that a thing so balanced as to stand in any position 
is not necessarily balanced for running : that a 4-lb. weight 
at 3 in. from the axis of rotation though balanced statically 
by a 1 lb. weight at 12 inches from the axis is not balanced 
by it dynamically. On the contrary, a 4-lb. weight at 5 in. 
is balanced by a 1-lb. weight at 10 in. from the axis." 



CHAPTER VIII. 
ELASTIC SOLIDS. 

160. Ix laying down the foundations of the subject it 
was stated as the result of observation and experiment 
that forces acting on a body might change its motion or its 
form, or that both changes might take place. The effect of 
the force depends on the nature of the body acted on. As 
most simple, we have considered first of all changes of mo- 
tion only. The action of the external forces was conceived 
to be resisted by the internal reactions of the particles on 
one another in such a way that the particles retained their 
original distances from one another so that change of form 
did not take place, or the body acted on was rigid. The 
conditions of equilibrium and of change of motion on this 
hypothesis have been developed in the preceding chapters. 

Experience shows that no perfectly rigid body exists in 
nature. The internal reactions do not prevent changes of 
form, the body yields to the external forces and changes its 
form or its size. If it returns towards its original configura- 
tion on the removal of the forces, it is said to be Elastic — a 
body possessing elasticity of form being called a solid, and 
one altogether devoid of elasticity of form a fluid. 

Conceive an elastic body (as a bent spring) in equilibrium 
under the action of given forces. When in this condition, 
nothing will be changed by supposing it to become rigid. 
The conditions of equilibrium of a rigid body may there- 
fore be applied to it in its distorted form, and the problem 
solved as for a rigid body. So that just as we use the par- 
otide as a stepping-stone to the rigid body, we use the rigid 
body as a stepping-stone to the elastic body. 

201 



202 



ELASTIC SOLIDS. 




161. The form of the body when in the position of 
equilibrium being distorted, differs from the original form. 
Hence, before we apply the conditions of equilibrium we 
must first of all inquire into the changes of form capable 

of being produced by 
the external forces. 
To do this it is neces- 
sary to appeal to ex- 
periment. 

Conceive a uniform 
beam with horizontal 
axis and resting on 
two supports under 
the action of external 
forces. These forces are transferred from particle to par- 
ticle of the beam, and give rise to internal forces or reac- 
tions. 

Let Fig. 168 represent a vertical section through the 
axis of the beam. For simplicity consider first the 
external forces to be parallel to this plane and let them 
be resolved into components along and at right angles 
to the axis. Conceive the beam divided, by a plane AB 
perpendicular to the axis, into two parts X and Y. The 
equilibrium will remain as before, provided forces equal 
and opposite to the internal forces at the section are 
applied. Let these forces be resolved into vertical and 
horizontal components, and let these be combined into 
single resultants, Q, Q; B, B. The forces of which Q, B 
are the resultants being distributed over the surfaces of 
the section and forming pairs in equilibrium, are known as 
Internal Stresses. 

The part X is in equilibrium under the action of the 
external forces and the stresses -f- B, — Q representing the 
action of J r upon X, and the part Y under the external 
forces and the stresses — B, -\- Q representing the action 



> 



STRESS AND STRAIN. 203 

of Xupon Y. In finding the relation between the exter- 
nal forces and the internal stresses the equilibrium of either 
X or Y may be considered. We shall take X, the section 
to the left of the cutting plane. 

162. To the stresses various names are given. Thus the 
stresses along the beam may tend to pull the pieces apart 
or to push them together. To the pull the name of tensile 
stress or Tension is given, and to the push the name of 
compressive stress or Compression. 

The transverse stress (consisting of equal and opposite 
forces -\- R, — R, considered acting indefinitely near the 
plane of section, but on opposite sides) causes the pieces to 
slide along the plane, and forms the shearing stress or Shear. 
The transverse stress may cause X, Y to turn about an 
axis in the plane AB, and forms the Bending Stress. 

If the external forces are not in the same plane, the 
bending stress may take place about an axis perpendicular 
to the plane AB, and forms the twisting or Torsion Stress. 
This generally occurs in machine shafting. 

These are the simple forms of stress. Usually stresses 
are compound, but may be resolved into two or more of the 
simple forms. 

163. When a beam is subjected to the action of a stress 
it yields. The change of dimension is known as a Strain. 
Remove the stress, and the beam returns to its original 
dimensions. This is observed to be true for stresses up to 
a certain amount. When that amount is exceeded, the 
beam will not return to its original form on removing the 
stress, but will assume another form between the two, or 
take a Permanent Set, as it is called. Increase the stress 
still further, and the beam will be finally ruptured. 

The discussion of stresses and strains forms the Mechan- 
ics of Materials, and will be found in special treatises. 



204 



ELASTIC SOLIDS. 



Fig. 170 




164. Impact. — Suppose a sphere of mass m to come in 
Fi I69 contact with another of massm,, 

a collision or Impact takes place. 
This impact is said to be direct 
if the bodies are moving in the 
same straight line, and the com- 
mon tangent plane at the point of 
meeting is perpendicular to the 
direction of motion: if not, the 
impact is said to be oblique. 

165. Direct Impact. — In conse- 
quence of the impact there is a 
mutual pressure produced, which 
increases the velocity of one of the 
bodies and diminishes that of the other. The velocity u 
of m is changed to v say, and the velocity u 1 of m, to v x . 
The impact action requires a certain time, which may be 
conceived divided into small intervals. Let P be the press- 
ure developed in any one of these intervals, and a, a x the 
accelerations produced in the masses m, m 1 , respectively. 
Then (Art. 34) P = ma, P — w?,«„ and 

. • . a : a , = m t : m. 

Similarly for all of the intervals. Hence, by addition, 

total accel. of m : total accel. of m 1 = m t : m, 

or v — u : u x — v t = m 1 : m, 

or mu + m x u x = mv -f- m l v l ; ... (1) 

that is, the sum of the momenta before impact is equal to 
the sum of the momenta after impact,— a statement indeed 
implied in the law of stress. 

This equation contains two unknowns, v and v t . We must 
therefore have another relation between them, as two equa- 
tions are necessary to determine two unknowns. Now it is 



IMPACT. 205 

found by experiment that when two bodies impinge direct- 
ly their relative velocity after impact bears a constant ratio 
to their relative velocity before impact so long as the 
material of the bodies is the same, but is in the opposite 
direction. This constant ratio is called the Coefficient of 
Restitution of the two bodies, ancf is denoted by the letter 
e. We have therefore 

v — i\= — e(u — wj .... (2) 

as the second relation between v and v x . Solving (1) and 
(2), we find the values of v and v x . 

The value of e depends on the material composing the 
bodies. From its definition it follows that the extreme 
values of e are and 1. If e == 0, or the bodies are In- 
elastic, then 

v — (mu + m ± a^/(m -j- mj ~ v xi . . (3) 

or the bodies move together with a common velocity after 
impact. If e = 1, then 

V — V x = — {% -r- Uj, 

or the velocity of one body relative to the other after im- 
pact is the same as it was before impact, but in the oppo- 
site direction. In this case the bodies are Elastic. 

No examples of either perfectly inelastic or of perfectly 
elastic bodies occur in nature. But some bodies with very 
little elasticity, as clay for example, may be regarded as be- 
longing to the first class, and others, as glass, to the second 
class. 

Ex. 1. Two inelastic balls are brought to rest by the im- 
pact: prove that they must have been moving in opposite 
directions, with velocities inversely proportional to their 
masses. 

2. Two balls of equal mass are perfectly elastic : prove 
that after impact they will exchange velocities. 

3. A row of equal elastic balls are placed in contact in a 




206 ELASTIC SOLIDS. 

straight line. An equal ball impinges directly on them. 
Show that all will remain at rest but the last, which will fly 
off. Verify experimentally. 

4. Find the elasticity of two balls of masses m and M in 
order that if M impinges on m at rest it will itself be 
brought to rest. Ans. M/m. 

166. Oblique Impact of a Sphere against a Fixed Smooth 
Plane AB. — Let u be the ve- 
F ! 9 ' m locity before impact, and v the 

velocity after impact; 6 the in- 
clination of u to the normal, and 
6 the inclination of v. 

Resolve the velocities along and 
normal to the plane. The plane 
being smooth, it exerts a normal pressure only. Hence the 
impact may be considered direct, with velocity u cos a be- 
fore and v cos /3 after impact ; and 

.* . v cos /3 = — eu cos 6. 

Also, since the pressure is normal, the action along the 
plane is unchanged by the impact. Hence 

v sin /? = u sin 6, 

and v and are found. 

Ex. 1. What are the values of v and /? above? 

2. If the elasticity be perfect, show that the angle of in- 
cidence 6 is equal to the angle of reflection /3. 

3. To hit a ball Q by a ball P after reflection from the 
edge CA of a billiard table. " Aim at a point B as far be- 
hind the edge CA as Q is in front of it." Prove this. 

4. Deduce a rule for reflection at two edges of the table. 

5. A ball impinges on an equal ball at rest at an angle 
of 45° to the line of impact: prove that if both are per- 
fectly elastic their velocities will be equal after impact. 

6. Two balls of given masses and moving in given direc- 
tions with given velocities impinge on one another: find 
the resultant velocity of each ball and its direction, 



IMPACT. 207 

167. When two bodies impinge, the Energy of Impact is 
broken into two parts, one being taken up in producing 
changes of form, heat, light, sound, etc., and the other the 
resultant motion of the bodies. With our usual notation 
we have for direct impact, 

energy before impact = (imi 2 -J- m x u*)/%, 

energy after impact = (mv* -j- mjo*)/% 9 

and the change during impact is therefore the difference 
between these two expressions. From Equations 1, 2, Art. 
165, we find by a simple reduction 

i(?nu* + m 1 u*)-i(mv*+m i v*)=Ul-& i ) m ™ x {u-uV, 

which gives the change of kinetic energy produced by the 
impact. The change is greatest when e = 0, or the bodies 
are inelastic. The expression being a positive quantity 
(since e 2 <1) would seem to indicate a loss of energy dur- 
ing impact. Whether the change is to be so regarded or 
not, depends on the end to be attained. If that is the 
propulsion of a missile or the driving of a pile, then change 
of form, heat, etc., are prejudicial, and the energy used in 
producing them is lost. If, on the other hand, change of 
-form is the main thing, as in moulding metal under a 
hammer or in riveting, this so-called loss becomes the use- 
ful energy, and the energy of motion useful in the former 
case becomes prejudicial in this. 

• Ex. Two trains of equal weight, moving with velocities of 
30 miles an hoar each and in opposite directions, collide: 
show that the loss of energy produced by the impact is the 
same as in the case of a train moving at 60 miles an hour 
striking another at rest. 

In the latter case find the velocity with which the debris 
will be moved along the track. 

Also, show that before impact the total energy in the one 
case is double that in the other. 



208 



ELASTIC SOLIDS. 



Fig. 172 



168. The case of impact that occurs most frequently in 
practice is when the bodies are inelastic, and one is at rest 
before impact. Placing e = 0, u i — 0, and substituting for 
v, v l their values from (3) Art. 165, we have for the col- 
lective energy of the two bodies 
before impact mu*/2, and after 
impact w?u*/2(m -f- m x ), making 
the change during impact 
mm 1 u i /2(m +wi,). 

Take for illustration the Pile 
Driver.* The principle is the 
same as in driving a nail with a 
hammer, except that the motion is 
always vertical, gravity being the 
force acting. 

When the ram of mass m in 
falling through a height h im- 
pinges on a pile of mass m % with 
velocity n, the pile is driven down- 
ward a certain distance s. Let 
F denote the resisting force of- 
fered by the ground. The work 
done on the resistance is Fs foot- 
poundals. The energy of impact 
is m 2 it 2 /2(m^\-7n 1 ) foot-poundals. 
The work done after impact by the force of gravity on ram 
and pile is (m -f m^gs foot-poundals. Hence 

m*u*/2(m + m t ) + [m + mjgs = Fs, 

and F is found. 

In gravitation units, if W = wt. of ram in pounds, w 
= weight of pile in pounds, and P = the resistance in 




* The figure represents a piledriver made by the Vulcan Iron 
Works, Chicago. 



IMPACT. 209 

pounds (or the ultimate load the pile will carry), then (re- 
membering that 2t 2 = 2gh) the above equation reduces to 

WVi/( W + w) + ( W + w)s = Ps, 

the standard form. 

At the last blow, the value of s being small, the second 
term may be disregarded in comparison with the first, and 
we have 

Wh/{W+w} = Ps. 

Still more approximately, by neglecting the weight of the 
pile in comparison with that of the ram, 

Wh = Ps. 

For example, to find the ultimate load a pile weighing 
500 lbs. could carry if the last blow from a height of 25 ft. 
of a one-ton ram sinks the pile one inch. The three form- 
ulas give 482,500; 480,000; 500,000 pounds, respectively. 

The Safe Load to be carried by the pile is some frac- 
tional part of this, which experience has determined. Thus 
with the second of the above formulas Col. Mason, U.S.A., 
in a series of experiments at Fort Montgomery, N. Y., 
found the fraction to be J, so that he proposed 

safe load = W*h/4t(W+ w)s; 

and Major Sanders, U.S.A., in an " extensive series of ex- 
periments made in establishing the foundations of Fort 
Delaware/' concluded that the third formula was to be 
depended on in the form 

safe load = Wh/Ss. 

The fact is, so much depends on local circumstances, 
particularly on the condition of the head of the pile at the 
last blow, that pile-driving formulas must remain essentially 
empirical. 



S1 ° BtASTIC SOLIDS. 

It is useful to notice that the energy of the ram before 

r in /( + W) " T al t0 nn, >' that is > *e greater 
»T is in comparison with w. Hence the ram should be 
large in weight compared with the pile. 

With the riveting hammer, steam hammer etc in 

w'^dorr th or, n ,s the e f to be attoined ' ^S-S 

3mm. • « y nei ' depends on Wwh/(W+w) 

which s the more nearly equal to Wh the greater W is in 
comparison with W, that is, the heavier the anvi I i n 
comparison with the hammer. 

ca,S- if' thrpile^iuS 0°t ft'at # 7^ 5 °° Ibs " to 

ft ^ H the 1 ,' •s! ,a ,r m "' " Cighi ^ 50 ° «"■ '-" st oke" "5 3 

q a -l • i- ^ w *- 270,000 foot-ixniiirh 

energy of recoil is gw?v*/%L at the 

most cases the time of impact is so small thalt is i, 
possible to measure it. If it oan be meagnred m ™_ 



IMPACT. 211 

mated, the acting force F is at once found from the rela- 
tion (Art. 34) 

Ft = mv. 

Thus suppose a hammer weighing 4 lbs. to strike a blow 
with a velocity of 16 ft. per sec, and that the time elapsiug 
from the first contact to the destruction of the motion is 
1/1000 second, the average force F acting for this time 
would be given by 

F= 4 X 16 X 1000 + 32 = 2000 pounds. 

The effect produced results from the enormous force 
developed in the short time. If the time be increased the 
force is diminished, in the same proportion. From this 
circumstance many familiar phenomena may be easily ac- 
counted for. 

Ex. 1. Why do we receive a severe jar from a step down- 
ward when one expects to step on the level ? 

2. Account for the different effects of a cannon-ball in 
striking a granite wall and an earth wall. 

3. The ram of a pile-driver of 250 lbs. falls through 10 
ft. and is stopped in -^ second : find the average force ex- 
erted. 

4. The head of a steam hammer weighs 3 tons. If steam 
is admitted on the under side only for lifting, and the 
head has a drop of 4 ft., find the average compressive force 
exerted during a blow if its duration is ^ second. 

Aas. 15 tons. 



CHAPTER IX. 
STATICS OF FLUIDS (HYDROSTATICS). 

170. According to the general scheme outlined in Art. 4, 
we now pass on to discuss the action of force on bodies in 
the fluid state. Our ideas of a fluid are derived from com- 
mon experience. Such substances as water, oil, air, steam, 
we call fluids. The one characteristic property by which 
they are distinguished from solids is the more or less ease 
with which the particles move among themselves; or in 
other words, fluids have less elasticity of form than solids. 
Some fluids offer more resistance to separation of the parti- 
cles from one another, or to flow, than do others — thus mo- 
lasses more than water, and water more than ether. This 
resistance to flow is known as Viscosity. We may, however, 
conceive of a fluid of such a kind as would offer no resist- 
ance whatever in any direction, the elasticity of form being- 
altogether wanting. A fluid possessing this property lias 
no existence in nature. It is an abstraction, just as a per- 
fectly rigid solid or a perfectly elastic solid is, and our only 
reason for introducing the idea is that it leads to simplicity 
of treatment. The more any fluid in nature differs from 
the hypothetical fluid, the greater the modifications neces- 
sary in the results derived from the hypothesis. As the 
basis of our reasoning, then, we lay down this definition : 

A fluid is a substance the particles of ivltich can be moved 
by any force hoivever small, and which act on one another or 
on any surface without friction. 

171. Two Kinds of Fluids.— It follows from the definition 
that a fluid offers no resistance to change of form. Hence 

212 



TWO KINDS OF FLUIDS. 213 

the fluid must be confined in a solid vessel or envelope either 
rigid or elastic. Suppose a fluid completely confined, and 
a force P applied to a piston entering the vessel. With 
some fluids (as water) the change of volume resulting from 
the force P will be so small as to be practically n il ; with 
others (as air) the change will be dependent on the force, 
and such that when the force is removed it returns to its 
former volume. A rough illustration is afforded by the ac- 
tion of a closed syringe and of a pop-gun. We therefore 
subdivide fluids into non-compressible or inelastic and 
compressible or elastic. The former are known as Liquids, 
the latter as Gases. We shall assume the absolute 
incompressibility of liquids and the perfect elasticity of 
gases, which assumptions though not strictly true are very 
nearly so, as repeated experiments have shown. 

172. The results of the action of forces on fluids not 
equally applied over the Whole surface of the fluid is chauge 
of form or flow. If the forces acting over the whole surface 
balance, the fluid is in equilibrium. We are thus led to 
distinguish the statics of fluids and the kinetics of fluids. 

The term Hydromechanics is used to designate the me- 
chanics of fluids. It is divided into Hydrostatics, which 
treats of fluids at rest ; and Hydrokinetics, which treats of 
fluids in motion. The term Pneumatics includes the prop- 
erties of gases as distinct from liquids. 

By some writers the term Hydraulics, originally applied 
to the motion of water through pipes, is taken to mean the 
mechanics of fluids. 

173. From the conception of perfect mobility among \\w 
particles of a fluid we infer that if a fluid is in equilibrium 
under the action of external forces — 

(a) Each particle must be equal?// pressed in every direc- 
tion. For if not, the particles would move in the direction 



214 STATICS OF FLUIDS. 

of least pressure, which contradicts the hypothesis of equi- 
librium : — 

(b) The force exerted by the fluid on any surface with 
which it is in contact (or the pressure on the surface) is nor- 
mal to the surface. For if not, friction must enter, which 
contradicts the hypothesis of perfect mobility. 

These are the two fundamental principles of the statics 
of fluids. 

174. Measurement of Pressure. — Suppose that a fluid is 
in equilibrium under the action of a number of external 
forces, and let P denote the pressure exerted by the fluid 
on an area A of the surface of the vessel containing it or of 
a surface immersed in it. Then P/A would be the press- 
ure exerted on a unit of surface if the pressure were uni- 
form over the surface, and would be the average pressure 
per unit if it were not uniform. Pressures are expressed 
in poundals per sq. in., pounds per sq. in., dynes per 
sq. cm., etc. 

By making the unit of surface indefinitely small, we have 
the pressure on an indefinitely small surface, and can there- 
fore speak in this sense of the pressure at a point. The 
expression is a conventional one. The pressure on a point 
is of course nil. 

175. Transmission of Pressure. — Since every particle of 
a fluid in equilibrium is pressed equally in all directions, any 
assigned particle must exert an equal pressure in all direc- 
tions on its adjoining particles, each of these an equal 
pressure on those adjoining, and so on throughout the fluid. 
Hence a pressure applied at any point is transmitted un- 
changed to every other point of the fluid, and a pressure 
applied to a surface is transmitted unchanged to every 
other unit of surface in contact with the fluid. This is 
sometimes called the principle of Pascal. 



TRANSMISSION OF PRESSURE. 



215 




Suppose a vessel full of water and furnished with two 
pistons fitting in two openings A 
and B. Let A contain A square 
inches of surface, and B contain 
B square inches. Suppose a force 
of P poundals applied at A bal- 
anced by a force of Q poundals 
applied at B. The pressure per 
sq. in. at A is P/A, and at B is 
Q/B. Hence 

P/A = Q/B, 

from which relation, when any 

three of the quantities P, Q, A, B are given, the fourth can 

be found. Experiment confirms this result. 

Again, suppose a plane CD to divide the fluid into two parts. 
The pressures on the two sides of this plane are equal, and 
are normal to the plane. There is equilibrium between 
the pressure on the surface CED and the pressure on one 
side of CD, also between the pressure on CAD and the 
pressure on the other side of CD. But CD is the projection of 
CED or of CAD, no matter what the form of these surfaces. 
Hence the pressure on a surface in a given direction is equal 
to the pressure on the projection of this surface on a plane 
normal to the given direction. 



Ex. Show that the pressure on a pump plunger is the 
same whether the end of the plunger is rounded or flat. 



Application: (1) The safety-valve of a water or steam 
engine. 

The pressure on the valve will show the pressure in the 
boiler, and by suitably placing the weight on the lever, this 
pressure may be adjusted to any amount desired. When 
it exceeds this amount the valve will be lifted and the steam 
escape. 



216 




Let TT,= wt. of valve in lbs., A its area in sq. in., Z,= dis- 
tance of centre of valve from fulcrum F in 
inches ; 
}V n = wt. of lever; / 2 = clist. of C. of G. of lever from 

fulcrum ; 
\y — wt. on lever; / 3 = clist. of centre of weight from 
fulcrum ; 
p — pressure in pounds per sq. in. of blowing-off. 
Take moments about the fulcrum, and 

toJt + 'wJt + w&^ip^A, 

the relation required. 

Ex. 1. The valve weight is 3 lbs., diameter of valve 4 in., 
distance from fulcrum to centre of weight 36 in., distance 
from fulcrum to centre of valve 4 in., weight of lever 7 
lbs., distance from fulcrum to centre of gravity of lever 15.5 
in. : find the weight at the end of the lever to make the 
blowing-off pressure 80 pounds per sq. in. 

An*. 108.3 lbs. 

2. With the same data find how far the weight must be 



TROTSMISSIOU OF PRESSURE. 



217 



placed from the fulcrum to make the blowing-off pressure 
75 pounds per sq. iu. Ans. 33.7 in. 



-The figure represents a section of 

w 



Fig. 175 



CHARGING SCREW 



(2) The IAfting-jach.- 

a Tangye jack. Water 
is forced from the cis- 
tern by the force-pump 
CG under the ram which 
works in a water-tight 
collar H. The weight 
to be raised is placed on 
the ram at W. 

By unscrewing the 
" lowering screw" the 
water is returned to the 
cistern. By means of 
the "air-screw" air is 
admitted when the jack 
is in use. 

Ex. 1. In a jack the 

plunger is 1 in. diameter 
and the ram 10 in. diam- 
eter. A weight of 10 
tons is to be raised : what 
pressure must be applied at the end of the lever, the lever- 
age being as 10 to 1. Ans. 20 pounds. 

2. In a jack the leverage is a : b, the pressure applied at 
the end of the lever P pounds, and the weight on the ram 
W lbs. Compare the diameters D, d of ram and plunger. 

Ans. D/d= \ f ~WbfP7i. 

176. It is evident from the above apparatus that water 
may be used to transmit force. Not only so, but it may 
be used for the storing of energy. For suppose the ram 
fixed, then the pressure by the action of the ram will 
remain stored until the ram is freed, when it will rise to 
the same height that it would have done if not interfered 




218 



STATICS OF FLUIDS. 



with. A vessel for the storing of energy in water is known 

as a Hydraulic Accumulator. 

The mode of action of an accumulator may be more evi- 
dent from Fig. 176. A is a 
solid ram working water-tight 
in a vertical cylinder, and car- 
rying a heavy weight ) I '. This 
weight, which in an accumu- 
lator usually consists of cast- 
iron blocks, is carried by a 
platform B, which latter is 
supported by rods from a 
cross-piece C, fastened to the 
top of the ram A. D and D 
are vertical guides. 

Water is driven into the 
chamber E through the pipe 
i^by means of a force-pump, 
thus raising the ram. The 



mechanism to be operated is connected with the pipe G, 
through which the energy stored is communicated. In this 
way hydraulic cranes, elevators, etc., may be worked from 
a central source by means of water-pressure. 

Ex. 1. The ram is 21 in. in diameter and the load 120 
tons : show that the water-pressure per sq. in. in the accumu- 
lator is 700 pounds, nearly. 

2. In the hydraulic machinery for opening and closing 
the lock-gates of the St. Mary's Falls Canal, at the entrance 
to Lake Superior, the ram of the accumulator has a diame- 
ter of 21 in., and carries a load of 20.76 tons. The water- 
engine driven by this accumulator makes one revolution 
per minute, the diameter of the piston is 15 in., and the 
length of stroke 96 in. Find the H. P. developed. 

Ans. 5.1, nearly. 

177. In the last two sections we have considered the trans- 
mission of pressure in vessels completely filled with fluid 




PRESSURES AT DIFFERENT DEPTHS. 219 

and acted on by external forces. We now proceed to con- 
sider the influence of the weight of the fluid on the surface 
pressed. 

178. Surface of a Liquid at Rest. — Consider a liquid in a 
vessel with the upper surface free and acted on by gravity 
only. This free surface is a horizontal plane. For the 
force of gravity on each particle is vertical, and as friction 
is wanting, the surface must arrange itself at right angles 
to the pressures, — that is, horizontally, — otherwise the parti- 
cles would glide over one another. 

This is the characteristic property of liquids as distin- 
guished from gases. ' The surface is an example of an equi- 
potential surface (Art. 143). 

179. Pressures at Different Depths. — Let (9 be a point in 
the liquid at a depth h below the surface. F! 9- l77 
Suppose a small horizontal circle ab of _^c a 
area A described about' as centre, and 
conceive the liquid contained in the verti- 
cal cylinder described on ab as base and 
extending to the surface cd to become 
solid. The equilibrium existing will not be disturbed 
thereby. 

The pressures on the sides of the cylinder being normal 
to the curved surface, are horizontal. Eesolving vertically, 
we have 

press, on ab — wt. (grav. force) of cylr. ad. 
Let p = press, per unit area of ab, d = mass of cubic unit 
of the liquid or its density, and this becomes 
];Xi = SAlig poundals, 
or p = gdh poundals 

= dh pounds. 
Thus the pressure varies directly as the depth beloic the 
surf ace of the liquid. 






220 STATICS OF FLUIDS. 

If the base ah is not horizontal the pressure will vary 
from point to point of the base, and p becomes the average 
pressure per unit surface. 

If we conceive the base to become smaller and smaller 
until it becomes indefinitely small, then p will represent 
the pressure at the point 0. Hence the pressure at any 
point within a liquid is proportional to the depth of the 
point beloiv the free surface of the liquid. 

The weight (mass) of a cubic foot of water at the tem- 
perature 4° 0. is 62.424 lbs., or 1000 oz., nearly. A cubic 
inch of mercury weighs 3429.5 grains, a cubic ft. 13,000 oz. 

Ex. 1. Find the pressure at a depth of 100 ft. in Lake 
Superior. Ans. 43.3 pounds per sq. in. 

2. A vessel is filled with mercury. At what depth is the 
pressure 20 pounds per sq. in. ? 

Ans. 20 = hx 3429.5 X 7000. 

180. Total Pressure on a Surface immersed. — This fol- 
Fig. 178 lows from the preceding by adding the 

pressures on all the unit surfaces contained 
in the given surface. Thus, if h x , h 2 . . . h n 

are the depths of the unit surfaces, the 
total pressure 

= g*K + g$h + • • • + fffiK 
= g${K + \ + . • • + K). 

Then, as in Art. 90, if A is the number of units of area 
in the surface and h the depth of its centre of gravity, 

X=(ixA l + ixA+...+ix K)/A 9 

or Ah = h l -\-h i -\- , . . + h n , 

and the total pressure on the surface =gdA h. 

That is, the total pressure on a surface immersed is equal 
to the weight (= gravity force) of a column of liquid whose 
lase is the area pressed, and height the depth of the centre 
of gravity beloiv the liquid level. 




TOTAL PRESSURE 0^ A SURFACE IMMERSED. 221 

Notice (a) that in this discussion nothing is said about 
the shape of the vessel. The result does not therefore de- 
pend on the shape. Also the surface pressed on may be 
placed anywhere in the vessel. It may therefore form its 
base. Hence the pressure on the base of a vessel contain- 
ing a liquid is independent of the shape of the vessel For 
example, the pressure against a dock would be the same if 
the dock were exposed to the Atlantic Ocean or were situ- 
ated in a land-locked harbor, provided the depth of water 
at the dock were the same in both cases and the water at 
rest. 

Ex. Explain the paradox, " Any quantity of liquid, how- 
ever small, may be made to support any quantity, however 
large." 

Notice (b) that a" pressure P on a surface A may be con- 
sidered as arising from a column of liquid whose base is 
the surface and whose height h is found from P = gdAh. 
This height is called the Head of Water. 

For a base of one sq. in. and head h ft. the pressure 
= 62.424A/144 = 0.434 h pounds ; and conversely, a pres- 
sure of p pounds per sq. in. corresponds to a head of 2.304 p 
feet. 

Ex. 1. Prove that the water pressure on a surface 1 ft. 
wide, h ft. deej), is 31^ h 2 pounds. 

2. Find the resultant pressure on a sluice gate, the water 
standing 10 ft. on one side and 6 ft. on the other. 

Ans. 1 ton. 

3. Compare the pressures on the upper and lower halves 
of a sluice gate. Ans. 1 to 3. 

4. The slope of a reservoir wall is 1 in 8 and the height 
25 ft. If the water reaches to the top of the sloping face 
find the horizontal pressure. Ans. 19,500 pounds. 

5. If two liquids which do not mix are placed in a bent 
tube open at the ends, prove that their respective heights 
are inversely as their weights. 

Hence, by attaching a graduated scale, show how to fiud 



222 STATICS OF FLUIDS. 

the relative weights of two liquids. Try water and mercury 
and see if the relation is 1 : 13. G. 

6. A sphere is filled with liquid. Account for the total 
pressure of the liquid on the surface being greater than the 
weight of the liquid. 

What is the relation between the pressure and the 
weight ? 

181. Centre of Pressure on a Plane Surface Immersed. — 
Suppose the plane surface of area A divided into areas of 
one unit each, and let h l9 # 2 , ... h n denote their respect- 
ive depths. The pressures on these areas are r/dh^ , gd/i 2 , 
. . . (jSh n , aud being all normal to the surface, and there- 
fore parallel, may be combined into a single resultant press- 
ure equal to their sum. The point in which this resultant 
meets the surface is called the Centre of Pressure. To find 
its depth h below the surface of the liquid proceed as in 
Art. 90, and 
h, = (ffSk l xh l +ff6h 9 Xh a + ... 

+g9K x K)/(f*h l + g*K+ • • - + ****)• 

Let 8 denote the inclination of the plane surface to the 
surface of the liquid, r l9 r a , . . . r n the distances of the 
unit jireas from the line of intersection of the two surfaces, 
and r the distance of the centre of gravity from this line ; 
then, since h 1 = r. sin 6, . . . , we have 

K = (>\ 2 + r': + . . . + r n 2 )/(>\_+ r - + • • • + '») 
= 0\ a + ?, 2 2 + . - . + r n *)/Ar (from Art. 90), 

= mom. inertia/area surf. X dist. of C. of G., 
a convenient formula. 

A case of common occurrence is a canal lock-gate stand- 
ing vertically. Suppose it to form a rectangle of breadth a, 
depth h, and with the upper edge in the liquid surface. 
Then 

h o = \W/hh X \U = \h 9 

or the point of application of the resultant pressure is at f 
of the depth of the rectangle. 



CENTRE OF PRESSURE ON A PLANE SURFACE. 



223 




182. We can now discuss the Stability of a Wall sub- 
jected to water pressure on one of its faces, — as a reservoir 
wall, for example. 

Suppose the cross-section of the wall rectangular or trape- 
zoidal, AB=a, DC =b, 
the face exposed to the 
water vertical, the 
height h, and that the 
water reaches to the top 
of the wall. 

The forces acting on 
the wall are the water 
pressure, the weight of 
the wall and the reac- 
tion of the ground sup- 
port. The resultant water pressure P on one foot length 
of wall = gdh x 7i/3, and acts horizontally through H when 
HC = h/3. The weight W of one foot length of wall 
= ^(a + V)ligS x , and acts vertically through G, the centre of 
gravity of the cross-section, S i being the density of the wall. 
Let the directions of P and W intersect in 0. Complete 
the parallelogram of forces to scale. The resultant R is 
equal and opposite to the reaction of the ground. Now, 
assuming that the wall is a single block, if the resultant R 
cuts the base between D and C the wall will stand ; if not, 
it will be overturned. (Art. 96.) 

The relation between the forces P and W for a certain 
assumed position of the resultant may be found by taking 
moments about the point in which its direction cuts CD. 
Thus if we assume R to pass through a point E we must 
have 



PX 0K= Wx EK 



as a condition to be satisfied, 



234 STATICS OF FLUIDS. 



The wall may also yield by sliding along its base The 
frictional force / between the wall and foundation is the 
product of the weight of the wall by the coefficient of fric- 
tion jj. between wall and foundation (Art. Ill), or 

Hence, assuming the wall to be incapable of sliding alono- 
any joint other than the base, we have, if the base is hori- 
zontal, and therefore P and /parallel, the condition 

to be satisfied when the wall is just on the point of sliding. 

Sometimes a wall is to be* designed which shall have a 
certain Factor of Safety, the meaning being that the thrust 
necessary to overturn the wall or to cause it to slide shall be 
an arbitrary multiple of the theoretical thrust. This mul- 
tiple is the factor of safety. 

Ex. 1. Which is at the greater depth in a rectangular area 
immersed m water, the centre of pressure or the centre of 
S ravit y ? Arts. Thee, of p. 

-, J' ^ he cross - section of an embankment which weighs 
125 lbs. per c. ft. is trapezoidal, with one face vertical. It 
is 5 It. wide at the top, 11 ft. wide at the bottom, and 15 ft 
high. Ihe water is to press against the vertical side, reach- 
ing to its top. Will the embankment stand ? Ans. Yes. 

3. The depth of thee, of p. of a vertical right-angled tri- 
angular lamina whose base lies in the surface is i the alti- 
tude: if the vertex is in the surface and base horizontal, it 
is f or the altitude. 

4. A reservoir wall, cross-section rectangular, height/; 
weight per cubic ft. tv 1 , water reaches to top of wall- find 
its thickness t that it may be on the point of being over- 
turned by the water pressure about the outer edge U. 

Ans. \wlf X -| = w x M X ~. 



UPWARD PRESSURE. 



225 



5. In (4) find the thickness of the wall that it may be 
just on the point of sliding on the base CD, the material 
weighing 125 lbs. per cubic ft. , and }x — 2/3. 

Ans. t = 0.387*. 

6. Show that whether a wall of rectangular cross-section 
is more likely to yield by rotation or by sliding depends 
on the coefficient of friction. 

7. The height of water on one side of a canal lock is a 
ft., and on the other side b 



Fig. 180 




ft. Show that the resultant 
pressure acts at a height 
(rt 2 + «6 + 6 2 )/3(« + 6) ft. 

8. The figure represents 
the cross-section of a dam 120 
ft. high. The masonry is 
divided into three principal 
sections. It is required to 
find where the resultant 
pressures cut CD( = 16 ft.), 
BF{=60 ft.), GH(=100 ft.). 
The dimensions are marked 
in the figure. The stone 
weighs 144 lbs. per cubic ft., 
and the water is on the left < 
face. 

Ans. Da = 3.9 ft., Fb = 22,1 ft., He = 43.8 ft. 

183. Upward Pressure. — Conceive a portion ABC of a 
fluid at rest to become solid. The equilib- 
rium will remain undisturbed. The forces 
acting on the solid portion are its weight 
vertically downward through its centre of 
gravity, and the fluid pressures normal to the 
surface at every point. The resultant of 
these pressures balances the weight, and must 
therefore act vertically upward through the 
0. of G. of ABC. 

Now if we place in the fluid a solid which displaces the 
same volume ABC, the upward pressure is the same as 



Fig. 181 




226 STATICS OF FLUIDS. 

before, because the conditions of equilibrium are the same; 
that is, the upward pressure on a solid which displaces a 
portion of fluid is equal in magnitude to the weight of the 
fluid displaced, and its direction passes through the centre 
of gravity of the fluid displaced. This is known as the 
principle of Archimedes* 

Ex. Infer from the principle of Archimedes that the 
loss of weight of a body immersed in a fluid is equal to the 
weight of the fluid displaced. 

184. A solid placed in a fluid will either float partially 
immersed or float wholly immersed, or sink. Two applica- 
tions of the principle of Archimedes to these cases are 
important. 

(1) Specific gravity ( = relative weight). Take a solid and 

p. |82 let its weight in air be W. Suspend 

it by a fine wire from a hook attached 

to one scale-pan of a pair of scales, 

and find its weight W x when immersed 

in a liquid. The loss of weight 

W — W } is the upward pressure, and 

is equal to the weight of an equal 

gpgl volume of the liquid. The ratio of 

i^gf the weight Wof the solid to the weight 

W — W 1 of the equal volume of liquid 

is called the specific gravity a of the solid with reference to 

the liquid taken. Thus 

ct= W/{W-W X ). 

* " Hiero, King of Syracuse, had a quantity of gold made into a 
crown, and suspecting that the workmen had abstracted some of the 
gold and used a portion, of alloy of the same weight in its place, 
applied to Archimedes to solve the difficulty. Archimedes, while re- 
flecting over this problem in his bath, observed the water running- 
over the sides of the bath, and it occurred to him that he was dis- 
placing a quantity of water equal to his own bulk, and therefore 
that a quantity of pure gold equal in weight to the crown would 
displace less water than the crown, the volume of any weight of alloy 
being greater than that of an equal weight of gold. It is related 
that he immediately ran out into the streets, crying out, ' evpr}Ka } 
GvpyKccl**' — J3esant, • m 




HYDROMETER. 227 

As in what is called "weighing" we really compare 
masses (Art. 39), so in specific gravity the idea is a relative 
weighing or a comparison of masses. The term may, 
therefore, be defined — 

Tlie specific gravity of a body is the ratio of the mass of 
the body to the mass of an equal volume of some standard 
body ; or since the density d of a body is the mass of nnit 
volume, it may be put in the equivalent form : the specific 
gravity of a body is the ratio of its density to the density' of 
some standard body. 

The standard taken is usually distilled water at its tem- 
perature of greatest density, 4° 0. This standard is arbi- 
trary, but is chosen on account of its convenience. 

Note that specific gravity does not depend on volume, but 
on equality of volume. As we may assume any volume of 
the standard, take a cubic unit of water (cu. ft. or cu. cm.). 
Denote its density by d. Then if <x be the sp. gr. of a body 
of density p 1} we have from the definition a = SJd, and 
the mass M of a volume U of the body is given by 

- M - U8 1 = Ucrd. 

Since 1 cubic ft. of water at 4° 0. weighs 62.4 lbs., then 

Weight or mass M in pounds = 62.4 U<t, 

Weight or pressure W in pounds = 62.4 Ucr, 

Weight or pressure W in poundals = 62.4 Ucrg, 

when U is expressed in cubic feet. 

Ex. 1. The true weight of a body is 25 grams. It weighs 
15 grams when immersed in water: find its specific gravity. 

Am. 2.5. 
*2. If W l} IF 2 , W 9 be the apparent weights of a body in 
three liquids the specific gravities of which are <j 1 , a 2 , a z , 
prove that 

WA«, - <r.) + h;k - <o + ";k - "J = o. 

3. A solid lighter than water, of weight IT, lias a sinker 
attached to it, and the two weigh W. when sunk in water. 



228 STATICS OF FLUIDS. 

The sinker alone weighs TF 2 in water: find the sp. gr. of the 
solid. Ans. W/(W+Wt-W,). 

Take a piece of cork with iron or lead for sinker. 

4. A man weighing 150 lbs. , with a cork life-preserver of 5 
lbs. attached, just floats in water. The sp. gr. of the cork 
being 0.25, find that of the man. Ans. 1.11. 

5. The sp. gr. of an iceberg is 0.925: find the ratio of 
the volume submerged to the volume exposed, the sp. gr. 
of sea water being 1.025. Ans. 37 to 4. 

6. A small stone of sp. gr. o~ is dropped into a lake of 
depth h: find the time in which it will reach the bottom. 

Ans. V'2licr/g{(T — 1) sec. 

7. A gold ring contains by weight 22 parts gold and 2 
parts copper : find the sp. gr. of the ring, that of gold being 
19.3 and of copper 8.9 . Ans. 17.6. 

Various modifications and extensions of the preceding 
method are in use. Thus to find the specific gravity of a 
liquid take a vial of weight W, fill it with liquid, and let the 
total weight be W x ; next fill it with water, and let the total 
weight be TF 2 : then 

a= (W-W)/{W-W). 

The same method may be used with a solid in fragments, 
or with soluble bodies, as sugar, salt, etc. 

185. (2) Again, for a solid lighter than water, if we float 
it in water and find the ratio of the volume immersed to the 
total volume, we have the valne of a according to the defini- 
tion. A convenient way of making the measurement is to 
use an upright cylinder of the solid graduated from bottom 
to top. Thus, if a cork cylinder 12 cm. high floats immersed 
to a depth of 3 cm., the sp. gr. of cork is 3/12 == 0.25. 

This same cylinder might be used to find the sp. gr. of a 
liquid in which it will float. The volumes displaced by the 
cylinder when placed iu water and in the liquid are ob- 
served; then 

a — vol. liq./vol, water = depth imm. in liq./do. in water. 



HYDROMETER. 229 

The common Hydrometer, invented by Boyle in 1675, is 
an instrument founded on this principle. It con- Fig. 183 
sists of a glass cylinder or rod arranged so as to 
float in a liquid in a vertical position. To effect 
this, two bulbs are added ; one, D, filled with air, 
and the terminal one, E, loaded with mercury or 
shot. Suppose that when the instrument floats in 
water the point G on the rod is in the surface, 
and when it floats in a liquid of sp. gr. cr the point 
B is in the surface. Let W — the wt. of the in- 
strument, U = its volume, and A the cross-section 
of the rod. Then 

W={U-A X OG)gS, 

W= {U-A X OB)gda; 

the weight of the instrument being equal to the 
weights of the liquids displaced. Hence 

U-A X OG= (U-A x OB)a, 

and the sp. gr. of the liquid is found. 

Hence we may graduate the rod so that when the instru- 
ment is placed in a liquid lighter than water the sp. gr. can 
be read off at sight. For by assigning different values to 
a in the above equation the corresponding values of OB 
are found, and may be marked on the rod. 

Ex. 1. How would you graduate the stem of a lactometer 
or milk hydrometer, so as to show various proportions of 
milk and water ? 

2. Explain the utility of a table of specific gravities. 

186. Floating Bodies. — In order that a solid may float 
in a liquid, it is necessary that the weight of the solid do 
not exceed that of the liquid displaced. If the floating 
body be slightly displaced from its position of equilibrium 
3,'nd it tends to return to that position, the equilibrium is 
said to be Stable ; if it tends to recede farther from that 



230 PNEUMATICS. 

position, the equilibrium is Unstable. The determination 
of the stability of floating bodies is of great importance in 
ship-building. The general problem is quite complicated. 
(See treatises on Naval Architecture.) 

Ex. 1. A segment of a solid sphere will float in stable 
equilibrium, the curved surface being down. Prove this. 

2. A prismatic block height a, breadth b, length /, and 
specific gravity a floats in water to a depth h : prove that 
equilibrium will be stable as long as b exceeds 1.225a. 

DYNAMICS OF GASES (PNEUMATICS). 

187. A perfect fluid (liquid or gas) is distinguished by 
the perfect mobility of its particles, elasticity of form or 
figure being altogether wanting. Hence all of the proper- 
ties regarding liquids, depending on this principle, which 
have been deduced in the preceding sections can be equally 
well applied to gases. The characteristic difference be- 
tween liquids and gases relates to behavior under pressure. 
Liquids are fluids with small compressibility, and gases 
fluids with large compressibility. The perfect liquid is 
taken to be absolutely incompressible, and the perfect gas 
to possess infinite compressibility. 

The gas with which we are most familiar and the one of 
most importance to us is atmospheric air. We live at the 
bottom of an ocean of air of whose existence we are con- 
scious from daily experience. Just as in describing liquids 
we took water as a near approximation to the perfect liquid, 
so we shall take air as an approximation to the perfect gas. 

188. Effect of Pressure. — Consider a portion of gas (air) 
enclosed in a cylinder and in equilibrium. On account of 
the perfect mobility among its particles, it will follow from 
the same reasoning as in Art. 175 that a pressure P ap- 
plied as by a piston will be equally transmitted in all direc- 
tions. The pressure at any point A within the cylinder is 
\\\q same in every direction, An illustration of the applica- 



BOYLE'S LAW. 231 

tion of equality of pressure of a gas in all directions has 
been already given in describing the safety-valve. 

If the pressure on the piston be increased the piston will 
descend in the cylinder, thus lessening the volume of the 
gas, but increasing its density. On removing the increased 
pressure the piston will return to its original position, 
showing that the gas possesses the power of expansion. 
Experiments with the ordinary gases go to show that with 
a perfect gas, if the pressure were doubled very slowly, the 
volume would be reduced to one half the original ; trebled, 
to one third the original ; and so on. In general, with a 
perfect gas the pressure varies inversely as the space occu- 
pied by the gas, the temperature remaining constant. This 
is known as Boyle's Law, and may be stated symbolically in 
either of the forms PU = c or P = c6, where P is the 
pressure, U the volume, S the density, and c a constant de- 
pending on the nature of the gas. 

A familiar illustration of the effect of pressure on the vol- 
ume of a gas is to take a tumbler and force its mouth down 
in a vessel of water. The farther down it is forced the 
greater the water pressure on the air inside and the less the 
volume becomes, as shown by the rise of water in the tum- 
bler. The mass of air is the same, no matter what the 
pressure, but the volume varies. 

189. Boyle's law may be represented graphically. Take 
OX, Y at right angles as axes of Y 
co-ordinates, and lay off OA to scale 
to represent U and A C to represent 
P : then (see any Analytical Geom- 
etry) the locus of a point C possess- 
ing the property OA X AC — a 
constant quantity, is a rectangular 
hyperbola, of which the axes OX, OY arc asymptotes. 
Hence this curve is a geometrical representation of Boyle's 
law. 




232 



PNEUMATICS. 



190. Extension to the Atmosphere. — We have so far con- 
fined ourselves to the properties of a gas when confined in 
a vessel. Boyle's law is founded on this limitation. So 
also are the principles of transmission of pressure, and of 
equality of pressure in all directions. In order to make 
use of these principles in the free atmosphere, we must 
assume that it is confined, and that the confiuing force is 
the force of gravity. The force of gravity keeps it encased 
in a huge envelope, as it were, surrounding the earth. 

191. Pressure of the Atmosphere. — Air being a material 
substance and subject to the action of gravity, must exert 
pressure downward as other material substances do ; that 
is, it must have weight.* Now the pressure of a fluid on 
any horizontal area is measured by the weight of a vertical 
column of the fluid whose base is the area. So the pressure 
of the atmosphere on any area would be measured by the 
weight of a column of air whose base is the area and height the 
height of the atmosphere. To measure this pressure take a 

bent tube of uniform bore (Fig. 185) 
and pour a quantity of mercury into 
it. The mercury will stand at the 
same height in both branches; or, in 
other words, the common surface AB 
will be a horizontal plane. The press- 
ures P, Q of the columns of air 
whose cross-sections are the areas of 
the tubes are equal, and therefore do 
not affect the level of the mercury in A and B. Suppose, 



Fig. 185 



*It is related that Aristotle (b.c. 384-322) "weighed a skiu first 
empty and then inflated with air, and finding it to weigh the same 
in both cases, concluded that air was without weight." The experi- 
ment failed because the weight of the skiu when inflated was dimin- 
ished by that of the equal volume of air which it displaced. It was 
reserved for Galileo (1564-1642) to show that air was a heavy fluid, 



PRESSURE OF THE ATMOSPHERE. 



233 



P = gdji, 



Fig. 186 



now, by pulling up a tight-fitting piston that the air is 

removed fro in B or the pressure Q 

removed, the mercury in B will rise 

and in A will fall until a certain 

difference of level h is reached, when 

it will remain stationary. Hence the 

weight of the column of mercury ~~ 

whose height is h and base B must be 

equal to the atmospheric pressure on 

the same area ; that is, 



when S x is the density of mercury. 

A simpler way of making this measurement is to take a 
Fig. 187 long glass tube of uniform bore and closed at 
one end, fill it with mercury, close the other end 
with the finger, invert the tube, and plunge this 
end into a basin of mercury. On removing the 
finger the mercury in the tube will sink until 
the height h is reached. This was the method 
of Torricelli,* in 1643, the first to make the ex- 
periment. 

The value of li is about 760 mm. or 30 in. 
Had the liquid used been water instead of mer- 
cury, the value of h would be 13.596 times 
greater, that is, 10,333 mm., or 33 feet. 

The atmospheric pressure per sq. in., p, fol- 
lows from considering that it is equivalent to the weight 
of a column of mercury of 1 sq. in. base and height 30 in. 



13.6 X 62.4 
Hence p = 32.2 X j=sg X 30 



473.3 poundals 



"A.d. 1G08-1G4T. Pupil of Galileo. 



234 PNEUMATICS. 

per sq. in. = 14.7 pounds per sq. in. This is sometimes 
called the pressure of one atmosphere. 

Ex. 1. The surface of an average man is about 6 sq. ft. 
Find the atmospheric pressure in tons. Ans. 6.4 tons. 

2. Show that the atmospheric pressure = 1,013,603 dynes 
per sq. cm. 

3. It is related that Otto von Guericke (a.d. 1002-1686), 
the inventor of the air-pump, in an exhibition before the 
Emperor Ferdinand III. placed side by side two hollow 
hemispheres of copper, and after exhausting the air showed 
that 30 horses, 15 back to back, were unable to pull them 
asunder. If d is the diameter of each hemisphere in 
inches and p the pull of each horse in pounds, show that/> 
did not exceed 0.1 7 cF. 

192. The instruments shown in Figs. 185, 187 may be 
used for finding the pressure of gases. If, for example, in 
Fig. 185, the tube A were connected with a steam-boiler, 
the heights to which the mercury rises in B will show the 
excess of the steam-pressure over the atmospheric pressure. 
This forms a siphon Manometer. 

The instrument shown in Fig. 187 forms a cistern Barom- 
eter, and will indicate atmospheric pressures according to 
the heights at which the mercury stands in the tube. For 
convenience of reading, a scale is usually attached to the 
tube. The term water barometer is applied to a barometer 
tube filled with water and of standard height, 33 ft.* 

Ex. 1. A cylindrical jar 2 ft. long is inverted and sunk in 
water until half-full : find the depth of its bottom below 
the surface of the water, the height of the water barometer 
being 33 ft. Ans. 32 ft. 

2. Show that the tension of the chain with which a div- 
ing-bell is lowered increases as the bell descends. 

3. A conical wine-glass is immersed mouth downwards 
in water : how far must it be depressed that the water within 
the glass may rise half-way up it ? Ans. 231ft. 

4. A cylindrical diving-bell of height I is sunk until its 
top is a foot below the surface: show that the height x of the 



PRESSURE OF THE ATMOSPHERE. 235 

air-space within the bell is found from x 2 -\- (a -j- h):c = hl 3 
where h is the height of the water barometer. 

5. In Ex. 4 find how much air must be forced into the 
bell to completely drive out the water. 

Ans. (a -f T)/li yolume of bell. 

193. Measurement of Elevations. — Since the pressure of 
the atmosphere at a place is equal to the weight of a column 
of air extending to the full height of the atmosphere at that 
place, it follows that the pressure will differ at different ele- 
vations, and that the height of the barometric column will 
give us a measure of this difference. We may therefore use 
the barometer as a measurer of differences of height. The 
first to propose this was Pascal in 1648. 

The solution of this question, with the precautions neces- 
sary for making a measurement, is quite complicated and 
beyond our province. 

194. Upward Pressure of 'the Atmosphere. — As in Art. 
183, we may show that the upward pressure of t±e atmos- 
phere on a body suspended in it is equal to the weight of the 
volume Z7of the air displaced, and that the direction of the 
pressure is vertical, and through the centre of gravity of the 
air volume. Hence if the weight of the body is less than that 
of the volume of air displaced, the body will float in the air. 
The balloon depends on this principle. 

True and Apparent Weight. — Since a body is pressed up- 
ward by a force equal to the weight of the air displaced by 
it, it follows that the apparent weight of a body (weight in 
air) is always less than its true weight (weight in a vacuum). 
The more bulky a body is the greater the difference. If we 
balance a bulky body, as cork, against a dense body, as lead, 
under the receiver of an air-pump and then exhaust the air, 
we shall find the cork descend. 

Ex. Show that the true weight of a ton of hay is more 
nearly found by weighing it in bales than in the loose form. 



236 



STATICS OF FLUIDS. 




195. Illustrations of Atmospheric Pressure. — (1) The 
siphon consists of a bent tube ABC 
open at both ends. It is filled with 
water and the ends inserted in two 
vessels of water as in the figure. 

If h is the height of the water ba- 
rometer and A the cross-section of the 
pipe, then the upward pressure at B 
in tube AB — gShA — gSA X BD, 
in tube BG — ydJiA — gSA X BE', 
. • . resultant pressure at B 

= gSA(BE - BD) = gSA X BE, 

and the water flows from A to C until BE = 0, or the sur- 
faces of the water in the two vessels are at the same level. 
An example of a siphon is seen in the common flush tank. 

Ex. 1. If C discharges into the air, explain Fig. 189 
the action. 

2. If the height of B above D is greater than 
the height of the water barometer, what would 
happen when the end A is opened ; when C is 
next opened ? 

3. The bore of a siphon is not uniform. Will 
the siphon work ? 

4. The apparatus in Fig. 188 is put under 
the receiver of an air-pump, and the air ex- 
hausted. What happens ? 

5. How would C need to be immersed to re- 
verse the flow ? 

6. Explain how ventilation may be effected 
by means of a tall chimney. 

(2) The common Lifting or Suction Pump 
consists of two tubes A, B, with the same axis, 
the lower terminating in the reservoir C; D 
is a movable piston operated by the rod E, and 
with a valve F in it opening upwards. Another valve G 



U=A_| 



LIFTING AND FORCE PUMPS. 



237 



opening upward is placed at the junction of the two cyl- 
inders. 

Suppose the pump filled with air and the piston D to 
be at its lowest limit. By raising the piston the valve F 
will close and the air in A will be carried with it. The 
air in B will open the valve G and rush into A. Hence at 
the surface of the water C the pressure inside the barrel is 
less than the atmospheric pressure and water will be forced 
up the tube C until equilibrium is restored. When the 
piston descends the air in A is compressed, G closes and 
F opens. Continue, and the water will rise until it pass 
through G, next through F } and will thence be lifted by the 
piston to the spout H. 

Ex. 1. Is it necessary to have two cylinders ? 

2. Suppose the tube B to be bent or irregular in figure, 
would the pump act? 

3. If the height of the valve G above the water surface 
is greater than the height of the water barometer, what will 
happen ? 

4. In a common pump the suction pipe is 10 
ft. long and the area of the barrel is 4 times that 
of the pipe. If the stroke is 2J ft., find how 
high the water will rise in the pipe at the first 
stroke. Ans. 7.5 ft., nearly. 

196. The Force-Pump differs from the lifting- 
pump in that the piston or plunger D is without a 
valve. On lowering the piston the air in the bar- 
rel is forced before it, the valve B opens, A shuts, 
and the air passes through B. On raising the 
piston the air in the barrel is carried with it, 
B is shut by the atmospheric pressure, and A is 
opened, allowing the atmospheric pressure to 
force water into the barrel. The piston in de- 
scending now forces the water through B as it 
did the air. 

The water will flow out in jerks, depending on the rapid- 



Fig. 190 



\ ' z 



238 



STATICS OF FLUIDS. 




ity of the up and down strokes of the piston. By placing 
an air chamber on the exit pipe the expansive force of the 
confined air in the upper part of this chamber will cause 
the stream to flow with a continuous but varying flow. 
The ordinary Syringe is an example of a force-pump. 
Fig. 191 The driving force is applied by squeezing 
the rubber ball A. The heart of an animal 
works as a syringe, the force being the mus- 
cular action of the walls. 

Ex. 1. The diameter of the plunger of a 
pump is 2 J in., and it is driven by a crank 
2 in. in length, making 30 revolutions per 
minute: find the number of gallons raised 
per hour. Ans. 153 gals. 

2. In a steam-pump the diameter of the 
water cylinder is 7 in., the length of stroke 
9 in., and the number of revolutions per minute 90 : find the 
discharge per minute. Ans. 135 gals. 

3. In a duplex pump the diameter of the water plunger 
is 7 in., the length of stroke 10 in., and the number of 
strokes of one plunger per minute is 100 : find the number 
of gallons delivered by the plunger in 1 min. 

Ans. 333.3 gals. 

197. The Air-Pump is a pump for removing air from 
a vessel. The common lifting-pump and the 
force-pump already described are air-pumps, as 
their first function is to cause a partial exhaus- 
tion of air before lifting or forcing the water. 
The labor of working an air-pump constructed 
as in Fig. 189 would be very great on account 
of the difference of pressure on the two sides of 
the piston. To relieve the piston a third valve 
is introduced as in Fig. 192. It is evident that 
with this pump the exhaustion can never be per- 
fect, as a certain amount of air must remain to 
lift the valves, which, even if made of oiled silk, as they com- 
monly are, still possess some weight. 



Fig 




THE AIR-PUMP. 



239 



Fig. 193 



K) 



A modification of the air force-pump, in which the solid 
piston is replaced by drops of mercury, is known as the 
Sprengel air-pump. It is used largely in the arts, as, for 
example, in exhausting the globes of incandescent lamps.* 
' A is the globe to be exhausted, B a basin of mercury, 
C a piece of tubing with a pinch-cock 
regulating the dropping of the mercury. 
A drop in falling acts as a solid piston, 
carrying with it the air in CD. The air 
in A expands and fills CD again. The 
next drop cuts this out, and so on, the 
air becoming rarer and rarer continu- 
ously. Almost perfect exhaustion can be 
had in this way. 

Ex. 1. If U l is the volume of the ves- 
sel to be exhausted by an air-pump, and 
U 2 the volume of the barrel of the pump, 
show that after two strokes the density 
tf 2 of the air in the vessel is found from 

where d is the density of the atmosphere. 

2. If the volume of the vessel to be exhausted is twice 
that of the pump barrel, find what proportion of the air 
remains at the end of three strokes. Ans. 8/27. 

3. If the volume of the vessel to be exhausted is twice 
that of the pump barrel, find after how many strokes the 
density of the air will be ^ of its original density. 

Ans. Between the fourth and fifth strokes. 

* The Berrenberg pump, a recent invention (1890), makes it "pos- 
sible to obtain rapidly and easily a vacuum comparable with that 
obtained with the best mercury pumps. The device is adopted of 
forming a vacuum jacket around the working cylinder so that what- 
ever leaks in around the outer side of the box, which is closed with 
oil as in ordinary air-pumps, is pumped out by the subsidiary cylin- 
ders. The difference of pressure therefore tending to cause leaks in 
the joints and valves of the working cylinder can be reduced to a 
quantity almost negligible.and the leakage almost disappears." — Elec- 
trical World, April, 1S90. 




240 



STATICS OF FLUIDS. 




198. Condensers. — If the tube B of a force-pump opens 
into a closed vessel A, then air may be driven 
into this vessel, and the apparatus becomes a 
condenser or conrpressor. 

The compressed air in the receiver may be 
called on to give up the energy stored in it by 
doing work, just as in the case of the water ac- 
cumulator. Air and water are thus of great 
use in transmitting energy in cases where 
gearing and belts are out of the question. 

In many cases air is more suitable than 
water for this purpose, as in the air-brake 
on locomotives, the pneumatic dispatch, the 
rock-drill, and more particularly in under- 
ground work, as mining, tunnelling, where 
the exhaust air is of use for ventilation. 

199. The exact determination of the work done in com- 
pressing air is a comj)licated question in thermodynamics, 
but we may make a first approximation by neglecting the 
changes of temperature undergone during compression. 
Suppose a volume U l of air under a pressure P y has its 
volume changed to U 2 by changing the pressure to P 2 . 
The relation between these quantities is given by Boyle's 
law to be 

PJJ X = PJJ„ 

each product being equal to a constant quantity. 

It was shown in Art. 189 that the geometrical representa- 
tion of Boyle's law is an equilat- 
eral hyperbola, of which the axes 
OX, OY are asymptotes. The 
ordinates of this curve represent 
the pressures, and the abscissas 
the corresponding volumes. 

Let OA = l\, AC=P } ; 
OB=U 2 , BD = P 2 . 



Fig. 195 




COXDENSEES. 241 

Now for an indefinitely small distance Aa along AB, that 
is, for an indefinitely small change of volume clU, the pres- 
sure P may be considered constant. Hence the work done 
in changing the volume from C T to U-\-dUis represented 
by P X clU, that is, by the element area AaCc. Sum- 
ming up the element areas from A to B, we find the total 
work done in compressing the gas from volume U 1 to vol- 
ume U 2 to be represented by the area ABDC. This area 
may be computed by Simpson's rule, or more simply by the 
aid of the polar planimeter, as in the case of the indicator 
diagram, Art. 136. 

Or it may be computed directly by summing the expres- 
sions P X dU between the limits U l and U a . We have 

work done ^ f' Pd U= F ' P Ud U/ U 

J T 2 U Da 

= P, U, log e XJJ l\ = P, T, log, PJP, ; 
or_changiug the logs to base 10, 

Work done = 2.3 P 1 U l log 10 P t /P l . 

If P x is in dynes and U 1 in cubic cm., the work done is 
in ergs ; if P l is in pounds and TJ X in cubic feet, the work 
done is in foot-pounds, etc. 

Ex. 1. Find the work necessary to compress 1 cubic foot 
of air to half its volume. 

Ans. 2.3 X (14.7 X 144) X log 10 2 = 1465.6 foot-pounds. 

2. Show that the work required to bring one cubic metre 
of air from atmospheric pressure to a pressure of S atmos- 
pheres is about 21,400 kilogrammetres. 

3. Find the H. P. required for a blowing engine to drive 
1000 cubic feet of air per minute from a pressure of 28 in. 
into a blast at a pressure of 30 in. Ans. 4 IT. P., nearly. 

4. How would you modify the apparatus in Pig. L81 to 
form a compressed-air manometer for measuring steam- 
pressures ? 



242 STATICS OF FLUIDS. 

5. Show how, from knowing the height to which water 
rises in a diving-bell, we can compute the depth of the bell. 

Hence show how a deep-sea-sounding apparatus might be 
made by sinking a tube closed at the upper end vertically 
into the s< 

200. The Work done by the Expansion of a Gas, as steam, 
for example, is calculated in the same way. 

Let P = the pressure of the steam on entering the cylin- 
der of an engine. After it has moved the piston through a 
distance I, let it be cut off and allowed to expand to the end 
of the stroke L. At any distance x between I and L the 
pressure ;; is found by Boyle's law from 

p : p = I : % or p = Pl/x. 

Hence for the work K done during expansion 

K =f L pdx = PI loge L/l = 2.3PI log 10 L/l 

per unit area of the piston. 

The work done by the steam up to the time of cut-off is 
PI X A if A is the piston area. Hence 

Total work = API(1 + 2.3 log 10 L/l). 

If A is expressed in sq. in., P in pounds per sq. in., the 
work done is in inch-pounds. 

Ex. 1. An engine with cylinder of 28 in. diameter and 
50 in. stroke carries 66 pounds of steam and is cut off at 
one-quarter stroke. Find the work done per stroke. 
Am. 616 X 66 X 12.5(1 + 2.3 log 4) = 1,213,000 inch-pounds. 

2. At n strokes per minute find the H. P. developed. 



CHAPTER X. 

KINETICS OF FLUIDS (HYDROKINETICS). 

201. The doctrine of the motion of fluids is called Hydro- 
kinetics. The subject is a very difficult one, and compara- 
tively little progress has been made in it. 

In order to discharge a fluid from a closed vessel con- 
taining it, an opening must be made in the vessel. If the 
opening is entirely submerged, it is called an Orifice ; if 
not, a Weir. The fluid will flow through this opening with 
a certain velocity v, and in a certain time t a certain 
quantity D will be discharged. The velocity, time, and 
quantity depend on circumstances which will form our 
first inquiry. 

If, in addition to a simple discharge, the fluid is to be 
conducted to another vessel as by pipes, the problem is 
complicated by our having to consider the mode of transit. 
This forms our second inquiry. 

Lastly, in its passage from one vessel to another, as the 
fluid possesses mass and velocity, and therefore energy, it 
may be made to overcome resistance and do work. This is 
our third inquiry. 

ORIFICES AND WEIRS. 

202. Velocity of Efflux.— To find the velocity of efflux u 
through an orifice, the depth of the Fi |96 

orifice being li. 

As the liquid issues from the 
orifice with the velocity v the sur- 
face in the vessel must sink with a 
certain velocity. To avoid consid- L 
ering this, we shall assume the ori- 

243 



A 


h 


A 


S33= 


§§ti 



244 



KINETICS OF FLUIDS. 



fice to be very small in comparison with the cross-section 
of the vessel. The height h may then be considered 
constant. Also, for the present the friction of the orifice 
or any other resistance that may enter will be neglected. 
We shall therefore find only what is termed the theoretical 
velocity of efflux. 

The pressure at the orifice is due to a column of the 
liquid of height li and cross-section A, and is equal to gdhA, 
which is therefore the effective force acting. The distance 
passed over by this column in a small time t is vt. 
Hence the work done during this passage by the column is 
gdhA X vt. 

Now the kinetic energy of the column issuing from the 
orifice with the velocity v is \SAvt X v 2 . Hence 



and 



gdhA X vt = idAvt X v* 

V = V2r/h 



= 8.02 \/h, nearly, — 



the velocity sought. This velocity is the same as that 
acquired by a body falling freely through a height h. (Art. 
57.) 



If the discharge takes place from one vessel to another, 
the effective force would be the 
difference of pressure at the two 
ends of the connecting tube, and 
therefore equal to gS Ah 1 —gd Ah ^ 
or to gdAli if h = h x — h 9 , and 
A is the area of the cross-section 
of the tube. Hence, as before, 
v 2 = 2gh when li is the effective 
head. 

If the water surface is subjected to pressure, we reduce 





Fig. 197 
* ft. 




SIS5S 


PSIS 


i§53|5Bi 




ISlljl 



TELOCITY OF EFFLUX. 245 

to a free surface by adding a column of weight equivalent 
to this pressure. Thus a pressure by a piston could be 
written in the form gdAh l9 that is, it could be replaced by 
a head of height li x , so that the total head to which the 
velocity of efflux is due would be h -j- h x , and 



v = ^2g(h + hJ. 

Again, if water flowed into a vessel with a velocity u, the 
head due to this velocity is u*/2g, and the total head 
of efflux is h -\- u 2 /2g. Hence 

v* = 2g (h + u*/2g) = 2gh + n\ 

Let A be the cross-section of the orifice, and A x that of 
the inflowing water: then if the inflow is equal to the out- 
flow, or uA — va, we have 



v = ¥2gh(l-A*/A l m ), 

and the velocity of outflow is determined. 

Ex. 1. Show that the head to which a velocity v is due 
is fouud from li — 0.016 v 2 . 

2. The piston of a force-pump is 9 in. in diameter, and 
it is driven down with a pressure of 2 tons: find the 
velocity of the issuing stream, neglecting all resistances. 

A /is. 97 ft. per sec. 

3. If the cross-section of the orifice is 0.1 that of the 
inflowing stream, show that the error in using the formula 
v = V'2gh to find the velocity of efflux is 0.5$. 

4. If A = A l} then v = x and u = x . Explain. 

5. The pressure in a boiler is 75 pounds per sq. in. 
Find the velocity of escape of the water into the air 
through a small orifice situated 1 ft. below the level of the 
water. 

203. The values found for velocity of discharge are 
theoretical, or such as would result did no resistances enter. 



246 



KINETICS OF FLUIDS. 



Fig. 198 



ill 



Ou account of such resistances the actual value of the ve- 
locity is smaller than the theoretical, and 
the relation between the two must be 
found by experiment. Observation shows 
that water flowing through an orifice 
with a sharp edge approaches the orifice 
in contracting lines. The minimum 
cross-section is at a short distance out- 
side the orifice, and this cross-section is 
about 0.64 that of the orifice. The velocity of the water 
at this section is about 0.97 of the theoretical velocity v. 
Hence, since theoretical discharge = cross-sec. A X vel. v, 
we have for the Actual Discharge D the expression 

D = 0.64 A X 0.97*; 

= OMAv. 

The number 0.62 is called the Coefficient of Discharge. 
Putting for v its value in terms of the head h, 

D = 0.62 J iftgh 

= 5 A tfh, 

which gives the actual discharge through a small orifice in 
a vessel. 

The experimental values here given are average values. 
They are found to vary with the head of water and with 
the nature of the orifice. For the coefficient of discharge 
we shall use the letter c. By putting c = 1 in the formu- 
las following the theoretic value is found, and by putting 
c = 0.62 the average actual value. 

Ex. An orifice of 1 sq. in. cross-section was found in 10 
min. to discharge 60 cu. ft. of water under a constant head 
of 9 ft. Compute the theoretic discharge, and find the co- 
efficient of contraction. 



CIRCULAR ORIFICES. 



247 




204. If the orifice lies in a vertical plane and is not small, 
the value of the head h is not the 
same for all parts of the orifice, and 
we must sum up from element to 
element in order to find the velocity 
of discharge. 

Suppose first the Orifice Circular. 
The summation is most readily ef- 
fected by using polar coordinates. 
Take the origin at 0, the centre of 
the orifice. Conceive the circle di- 
vided by concentric circles into rings 
of width dr, and let each ring be divided into elements PQ 
included between two radii inclined at an angle dd. Let 
r, 6 be the coordinates of P. Then area of element PQ — 
rdrdd. Let h = depth of below the surface, then depth 
of element PQ — li — r cos 6. Hence 

D =rj^ 2n crdrdd V'2g(h - r cos 6) 

= c nr* V2gh(l - r 2 /327* 2 ), nearly, 

= 15.63r 2 Vl{\ - r/32/* 2 ), approx. 

If the orifice reaches to the water surface, h — r, and 

D = 0.9646'7rr 2 Vf&gA 

= lo.llr VTi, approximately. 

Ex. 1. Water is discharged through a circular orifice of 
diameter d in. under a head of li ft. Show that the dis- 
charge in gallons per minute is 12. 2d 2 Vh, nearly. 

2. How many cubic feet per second would be discharged 
through an orifice 1 in. in diam. under a head of 25 ft. ? 

Ans. 0.135 c. ft. per sec. 

3. Find the diameter of an orifice that would discharge 



146 gals, per min. under u head of 9 ft.? 



Ans, 2 in. 



248 



KINETICS OF FLUIDS. 



Fig. 200 




and 



4. Find the head necessary to discharge 110 gals, per 
min. through a 3-in. circular orifice ? Ans. 1 ft. 

205. Suppose next the Orifice Rectangular, of breadth b, 
and the depth of its lower and upper 
edges below the water surface h i9 h 2 . 
Consider a strip at a depth y below the 
surface, and of indefinitely small width 
cly. Along this strip v = c V2gy, and the 
discharge through this strip —cb V'2gydy, 
Hence 

D ^Ju x ch ^9V d H 
= \cb V¥g(hl - hf) 
= 3.33£(A§ - 111), 

average vel. of efflux = D/b{li — /*,), 

= 3.33(7*1 - hf)/{h - h t ). 

If 7i i= 0, or the upper edge is in the surface, the opening 
becomes a slot of breadth b and Fig. 201 

depth h, and is known as an over- 
fall or Weir. Here^ 

D = \cbli Vtyh 

= 3.33M* cu. ft. per sec, 
and 
aver. vel. = 3.337*4 ft. per sec. 

Ex. 1. The top of a rectangular orifice in the side of a ves- 
sel is 2 ft. and the bottom 3 ft. below the surface: find the 
discharge and the mean velocity of efflux, the orifice being 
2 ft. wide. Ans. 1G cub. ft. per sec; 8 ft. per sec. 

2. Show that the number of gals, per min. that will pass 
through a rectangular notch of breadth b and depth h, both 
expressed in inches, is 3#Af . 

3. In a rectangular weir the head is 1 ft.: find the aver- 
age velocity. Ans, 3.33 ft. per sec 




TIME OF EFFLUX. 



249 




D ■ 



:2C 



206. Of late a weir in the form of a Triangle has been 
Fig. 202 ^_^ introduced as possessing certain advan- 
tages, notably simplicity, over the rectan- 
gular form. 

Let the figure represent the weir, the 
angle ABC = 90°, and AB = BO. Take 
h = OB, the head of water, and let OC, 
OB be the axes of x and y. Then 

2c f Q xdy V2j7j 

= 0.5S3¥cV2gh 

= 2.6/22 cu. ft. per sec 

Ex. In a right-triangular weir the head is 6 in. : find 
the discharge. Ans. 0.46 cu. ft. per sec. 

207. In California, water supplied for irrigation and other 
purposes is usually expressed in Miner's Inches, meaning 
the average flow through each-square inch of a rectangular 
orifice under a given head-.- The discharge per minute is 
the minute-inch, per hour the~hour-inch, etc. The term 
is not a very definite one, but experiments made to deter- 
mine it agree that a miner's inch will discharge 1.2 cubic 
ft. per min. under a 4-in. head. The miner's inch is thus 
equal to about 9 gals, per min. 

208. To find the Time of Efflux through a small orifice 
of cross-section A l in the bottom of 
q» vessel of cross-section A. 

In consequence of the efflux the 
surface A must sink with a certain 
velocity u. The volume of sinking 
must be equal to the volume of out- 
flow ; and therefore for the indefi- 
nitely small time dt, during which u 
and the velocity of efflux v may be 



Fig. 203 



Y 



250 



KIKETICS OF FLUIDS. 



considered constant, we have 

Au = A x v. 

Now during this same time h may be considered constant, 
and we may write 

v = c V2gh. 

Also dh is the distance sunk through in the time dt with 
the velocity u, so that dh — udt. Hence 



and 



Adh/dt = cA 1 V2gk, 



cA.Jh 



dh 



cAyh V2 gh 
= 2A \li/cA l V2g' 
= 2 A VJi/5A lf approx. 

Ex. In what time would a bath-tub supposed of uniform 
cross-section, 8 ft. long, 2 ft. wide, empty through an inch 
hole in the bottom, if the water stood at a depth of 2 ft. 

Ans. 2.75 min., nearly. 

FLOW FN - PIPES. 

209. Short Pipes. — Consider the usual case of water enter- 
ing a circular pipe with square edges at the entrance. If 
the pipe is short, or the length about 
two or three times the diameter, the 
velocity of outflow v 1 is found by exper- 
iment to be equal to c x v, when v is the 
theoretical velocity V2gh, and c, is a 
constant, depending on the head of 
water and on the size of the pipe. An 
average value of c 1 is 0. 82. 

We may therefore write 

v x - c x v - c x V2gh = 6.58 Vh, 



1 Fig. 204 







FLOW IN PIPES. 251 

and the discharge D through a pipe of diameter d is 
D = c x vx n&*/i = 5.15d 2 Vh. 

The resistance of the orifice at influx may be expressed as 
a " loss of head" h 3 . The total head h may thus be regarded 
as consisting of two parts, h x , 7i 2 , of which h x generates the 
velocity of efflux v lf and // 2 is spent in overcoming the 
resistance at entrance into the pipe. Now since 





V%gh x — v x = c x V%gh, 


we have 


K = c*h, 


and 


h 2 = % i — h 




= *,(! - l/c*) 




= cji x , suppose. 


Putting 


c x = 0.82, this gives 




\ = 0.67//, 




h 2 = 0.507i i =0.34//. 



The effective head is thus 2/3 of the total head, 1/3 being 
dissipated at entrance. 

Ex. 1. Find the quantity discharged in one min. by a 
short tube 3 in. diam. under a head of 4 ft. 

Ans, 38.6 cu. ft. 

2. Find the head under which 460 gals, per min. will be 
discharged by a short tube 5 in. diam. Ans, 1.2 ft. 

3. Show that the discharge through a short pipe = 
15.7<F Vh gals, per min., when // is expressed in ft. and d 
in inches. 

210. Long Pipes. — If the pipe is long, in addition to the 
loss of head // 2 arising from the resistance at entrance, there 
is an additional loss of head // 3 arising from the friction of 
the water in the pipe. The total loss of head is therefore 
//„ + K • 



252 KINETICS OF FLUIDS. 

The loss of head 7z 3 is found by experiment to be directly 
proportional to the length of the pipe /, and the velocity 
head of efflux \ , but inversely proportional to the diame- 
ter of the pipe. "We may therefore write 

h 3 = cjhjd, 
when c 3 is a constant. An average value of c ? is 1/40. 

Hence h, the total head of water which is considered as 
consumed in overcoming the resistance of entrance equal 
to a head of h 2 , the resistance of friction equal to a head 
of h s , and in producing a velocity v 1 , equal to a head of h x , 
may be written 

= % x + 0.6071, + O.OZmjd; 

and the velocity of efflux, 

v i = ^tyK 

= -|/2^/(1.50 + 0.025l/d). 

The values of v r for given lengths and diameters of pipe 
may be tabulated for convenience of computation. 

Approximately, in a very long pipe, the length exceeding 
1000 times the diameter, the head due to friction so much 
exceeds the others that it may be considered equal to the 
total head, so that 

h = OMUhJd = 0.025^72^7, 
and the velocity of discharge 

v x = 50 Vhd/l ft. per sec. 
Also, D=v } X 7td 2 /4: 

— 40 VluVfl c. ft. per sec; — 
convenient approximate rules. 



FLOW IX PIPES. 



253 



This may be put in still more simple shape by calling A 
the area of the cross-section of the pipe, i the slope or fall 
in feet per mile, whence h/l = {/52S0. And 

v = § Vdi 

D = §A Vdi ;— 

rules sufficient for first approximations. 

If a number of small vertical tubes of equal cross-section 
be inserted in a pipe through, which water is flowing, the 
water columns will stand at different heights in these tubes, 



Fig. 205 




showing the pressure heads at the points of insertion. The 
upper surfaces of these columns will lie in a straight line 
passing through the exit end of the pipe, and called the 
hydraulic grade-line. The tubes themselves are called 
piezometers. 

If a pipe is laid so that part of it is above the hydraulic 
grade-line, it will act as a siphon. 

Ex. 1. Find the discharge of a set of pipes 1000 ft. long 
and 6 in. diameter under a head of 5 ft. 

A us. 0.5 c. ft. per sec. 

2. A set of pipes 1000 ft. long is required to discharge 2 
cubic ft. per sec. under a head of 20 ft. : find their diameter. 

Ans. 8 in. 

3. With the same head and length, a 2-in. pipe will de- 
liver nearly 6 timesand a3-in. pipe nearly 16 times as much 
as a 1-in. pipe. 



254 KINETICS OF FLUIDS. 

4. A 6-in. pipe 2^ miles long discbarges 30 cubic ft. per 
minute : find the head. Ans. 66 ft. 

5. Find the H. P. necessary to raise 1,000,000 gals, of 
water in 24 hours to a height of 100 ft. through a pipe 
5000 ft. in length. 

211. In an Open Channel as a canal or stream the flow is 
subject to laws similar to those governing the flow in pipes. 
The consideration of this subject will be found in special 
treatises. 

ENERGY OF FLOW. 

212. The kinetic energy of a body of water of mass m 
and moving with a velocity v is %mv\ If D is the sup- 
ply in cubic ft. per sec. and h the head in ft. corresponding 
to the velocity v in ft. per sec, then the kinetic energy 
per sec. = {X 62.4 Dv 9 ft.-poundals = 62.4 Dli ft.- 
pounds (since v 1 = 2gh), and the horse-power of the stream 
= 62.4 ZM/550 = 0.1135 Dh. This energy may be utilized 
in various ways in the driving of hydraulic machines. 

It may also act destructively. For suppose water 
flowing through a pipe is suddenly checked by closing a 
valve, and the flow stopped. The energy of motion reacts 
on the valve and pipe, and imparts a shock known as the 
Water Ram. Thus in a pipe a mile long, 1 ft. in diameter, 
and in which water is flowing at 2£ m. an hour, the energy 

= 7f M) 2 X 5280 X 62.4 X (V) 2 ft.-poundals => 54,380 

ft. -pounds. This very large number will show the need of 
a large factor of safety in designing pipes. 

Ex. 1. Show that the energy of flow varies as the cube 
of the velocity, the cross-section being uniform. 

2. Show that the H. P. of a stream supplying W lbs. per 
sec. under a head of h ft. is 0.0018 Wh. 

3. Show that it takes about 6.5 cubic ft. per second under 
a head of 8 ft. to give one H. P. 

4. Find the H. P. of a waterfall 18 ft. high, if the 
stream at the fall passes through a notch 2 ft. X 3 ft. with 
a velocity of 2 J m. an hour. Ans. 45 H. P, 



ENEKGY OF FLOW. 255 

5. The cross-section of Niagara River is 40,000 sq. ft. 
and the current 7 m. an hour. Show that this current 
represents an energy of the water equivalent to a head of 
1.6 ft. 

6. In (5) show that if all of this head could be utilized, 
the energy available would be about 75,000 H. P. 

7. If the energy taken out amounted to reducing the 
current velocity from 7 miles to 6 miles an hour, instead of 
to rest as in (6), to what H. P. is this equivalent ? 

8. If a tunnel 20 ft. diameter and 10 miles long were dug 
from the Niagara River to Buffalo, with a fall of 8 ft. per 
mile, and were running constantly full, show that the H. P. 
developed would be about 30,000. 



INDEX. 



Absolute units, 41. 

Acceleration, 15, 17: unit of, 
15: curve of, 17: angular, 
181: central, 72 : tangential, 
181 : of point moving in a 
circle, 72. 

Accelerations, composition of, 
22: resolution of, 23. 

Accumulator, hydraulic, 218. 

Air pump, 238. 

Amplitude of S. H. M., 77. 

Angular acceleration, 181. 

Angular velocity, 179. 

Arm of couple, 96. 

Atmosphere, pressure of, 232. 

Attraction, 127. 

Atwood Machine, 59 

Axis, instantaneous, 176. 

Axle friction, 136. 

Balance, 112. 

Barometer, cistern, 234 : water, 

234. 
Beam, 111, 115. 
Bell crank lever, 100. 
Belt friction, 139. 
Belts, driving force of, 140. 
Bending stress, 203. 
Blackburn pendulum, 78. 
Boyle's law, 231. 

Capstan, 101. 

Central forces, 66. 

Centre, of parallel forces, 102: 
of gravity, 103: of oscilla- 
tion, 192: instantaneous, 176. 

Centrifugal force (so called), 
73: effect of, on weight, 75: 
effect of, on belt tension, 141. 

Centrifugal pendulum, 83. 



Centripetal force, 72. 

Centrode, 177. 

Centroid, 104. 

Circular motion, 72, 81. 

Coefficient, of inertia, 33: of 
friction, 130: of restitution, 
205. 

Components, 20, 46. 

Composition, of displacements, 
19 : velocities, 19 : accelera- 
tions, 22: S. H. ' M.'s, 78: 
forces on a particle, 44, 45, 
48, 51 : forces on a body, 86 : 
couples, 96. 

Compression, 31, 203. 

Condenser, 240. 

Conservation of energy, 169. 

Constrained motion, ""57, 68. 

Couple, arm of, 96. 

Couples, composition of, 96. 

Degrees of freedom, 57. 
Density, 105: unit of, 105. 
Differential pulley, 143: screw, 

145. 
Dimensions of units, 42. 
Discharge through an orifice, 

246-248. 
Displacement, 9. 
Dynamics, 30. 
Dyne, 35. 

Effective force, 182. 

Efficiency, 161. 

Efflux, velocity of, 243: time 
of, 249. 

Elastic bodies, 201. 

Energy, 164: principle o\' kine- 
tic, 165: potential, 108: law 
of conservation of, Hi!). 
857 



258 



INDEX. 



Energy of rotation, 197: of im- 
pact, 207: of efflux, 254. 

Equations of motion, of a par- 
ticle, 36: of a body, 181. 

Equilibrium, 53: of a particle, 
55, 151: of a rigid body, 98, 
153: of a system, 153. 

Equipotential surface,. 171. 

Erg, 150. 

Factor of safety, 224. 

Falling bodies, GO. 

Field of force. 170. 

Floating bodies, 229. 

Flow in pipes, 250. 

Fluid, 212: pressure of, 214, 
219, 225. 

Fly wheel, 198. 

Foot, 41. 

Foot-pound, 150. 

Foot-poundal, 150. 

Force, 7, 33: units of, 35, 39: 
representation of, 45: ele- 
ments of, 45: centripetal, 72: 
diagram, 48: moment of, 92. 

Forces, composition of, 44, 48, 
86, 90: resolution of, 50. 

Free motion, 57. 

Freedom, degrees of, 57. 

Friction, 129: coefficient of, 
130: angle of, 132: axle or 
journal, 136: belt, 139. 

Governor, engine, 85. 

Gram, 42. 

Graphic statics, of simple 

trusses, 120: of a mechanism, 

125 
Gravitation, law of, 127, 170: 

units of force, 39. 
Gravity, force of, 38: centre of, 

103: specific, 226. 
Gyration, radius of, 190. 

Harmonic motion (S.H.M), 76. 
Head of water, 221. 
Hodograph, 26. 
Hoisting machine, 154. 
Horse power, 159: of a steam 
engine, 163. 



Hydraulic, accumulator, 218: 

grade line, 253: jack, 217. 
Hydrokinetics, 243. 
Hydrometer, 229. 
Hydrostatics, 212. 

Impact, 204: energy of, 207. 

Impulse, 36: average force of, 
211. 

Inclined plane, motion on, 69: 
equilibrium on, 71. 

Indicator diagram, 158. 

Inertia, 32: law of, 33: coeffi- 
cient of, 33 : moment of, 
184. 

Instantaneous, axis, 176: centre, 
176. 

Jack, screw, 154: hydraulic, 

217. 
Jointed frames, 120. 
Joule, 150. 

Kilogram, 42. 
Kinematics, 8. 
Kinetic energy, 165: change of, 

on impact, 207. 
Kinetics, 54. 

Laws of motion, 32, 33, 36. 
Length, units of, 41. 
Level surface, 219. 
Lever, 99, 153. 
Long pipes, flow in, 251. 
Longitudinal stress, 119. 

Manometer, 234. 

Mass, 34: unit of, 35. 

Mass-acceleration, law of, 36. 

Mechanics defined, 7. 

Meter, 41. 

Miner's inch, 249. 

Moment, of a force, 92: unit of, 
93: graphical representation 
of, 94: of inertia, 184. 

Momentum, 36. 

Motion, in a st. line, 10: in a 
curve, 24: in a circle, 72, 81: 
harmonic, 76 : on a rough 
surface, 135: relative, 27. 



259 



Neutral equilibrium, 109. 
Newton's laws of motion, J 



Oblique impact, 206. 
Orifices and weirs, 243. 
Oscillation of a pendulum, 79, 
193. 



Parallel forces, 90. 

Parallelogram of displace- 
ments, 19: accelerations, 19: 
velocities, 22: forces, 46. 

Particle, 8. 

Path, 10. 

Pendulum, simple, 77 : Black- 
burn, 78 : centrifugal, 83 : 
physical, 191. 

Period of S. H. M., 77. 

Permanent set, 203. 

Phase of S. H. M., 77. 

Pile driver, 208. 

Planetary motion, 68. 

Pneumatics, 230. 

Pole of stress diagram, 88. 

Polygon of forces. 55. 

Position of a force, 88. 

Potential, 169. 

Potential energy, 108. 

Pound, 35, 39, 42. 

Poundal, 35. 

Power, 158: units of, 159. 

Projectiles, 62. 

Prony brake, 160. 

Pulley, 91: differential, 143: 
tackle. 142. 

Pump, Lifting, 236. Force, 237: 
Air, 238. 



Radius of gyration, 190. 
Range of a projectile, 64. 
Relative motion, 27. 
Repose, angle of, 133. 
Resolution, of accelerations, 23: 

of forces, 50. 
Resultant, 45, 48, 51. 
Rigid bodies, 174. 
Rotation, 175: energy of, 197. 



Safety valve, 215. 

Screw, pitch of, 145: efficiency, 
147: differential, 145: jack, 
154. 

Shear, 203. 

Siphon, 236. 

Smooth surface, 129. 

Specific gravity, 226. 

Speed, 11. 

Sprengel air pump, 239. 

Stability, 108: of a retaining 
wall, 110: of a reservoir wall, 
223. 

Standards of length, 41. 

Statics, 53. 

Steam engine, 13, 17, 125, 194. 

Steam hammer, 210. 

Steelyard, 114. 

Strain, 203. 

Stress, 31: law of, 32: longi- 
tudinal and transverse, 119: 
diagram of, 122: shearing. 
203: in framework, 120: in 
steam engine, 125. 

Strut, 122. 

Syringe, 238. 

Tension, 31, 203. 

Thread of screw, 145. 

Tie, 122. 

Time, unit of, 41: of efflux, 

249. 
Toggle joint, 196. 
Torque, 96. 
Torsion stress, 203. 
Translation, 19, 44. 
Transmission of fluid pressure, 

214. 
Triangle^of forces, 54. 

Units, fundamental, 41: of 

work, 150: of power, 159: of 

angular velocity, 179: of 
angular acceleration, 181. 

Velocities, composition of, '21: 

resolution of, 23. 
Velocity, 10, 14: unit of. 11: 

curve of, 12: angular, 179. 
Viscosity, 212. 



2G0 



INDEX. 



Water ram, 254. 

Watt, the, 159. 

Weight. 40: effect of centrif- 
ugal force on, 75: true and 
apparent, 235. 

Weirs, 248. 

Winch, 102, 155. 



Work done, 148: unit of, 150: 
dimensions of, 150: principle 
of, 152: against friction, 156: 
against a variable force, 157: 
in compressing air, 240: by 
expansion of a gas, 242. 

Worm wheel, 155. 



